Just by chance I found out that (CH) implies $\omega_2 = \omega_2^\omega$, which I found quite surprising, since this an almost trivial conclusion of
$2^{\omega_1} = \omega_2$ rather than of $2^{\omega} = \omega_1$.
So, I'm wondering, if this implication is known, and if yes, I would appreciate to get a reference for this.
BTW: I'm always working in (ZFC). I think in particular the above mentioned proof uses a lot of choice.
(CH) implies $\omega_2 = \omega_2^\omega$: reference
cardinalsset-theory
Related Solutions
Edit: I have added an argument that shows choice in $M$ in terms of existence of choice functions.
Replacement in M: Here is an argument that shows $ZF^-$ in $M$ (the same works for $N$ of course):
Let $\mathbb R^G$ denote $\mathbb R^{L[G]}$. In a first step, we will see that in $M$ any definable map $f:\mathbb R^G\rightarrow \omega_2$ is bounded. In a second step I will explain how replacement follows. In $M$, all sets are definable from ordinal parameters, reals and proper initial segments of $G$. Since any real and any proper inital segment of $G$ is in an intermediate extension $L[G\upharpoonright \alpha]$ and since $L[G]$ is a further $Add(\omega, \omega_2)$-extension from there, we might as well assume that all parameters in the definition of $f$ are from the ground model and will suppress them in what follows. If $\dot x$ is a $Add(\omega, \omega)$-name, then by ccc, there is $\alpha_{\dot x}<\omega_2$ so that $1\Vdash f^{L_{\check\omega_2}(\mathbb R^\dot G)[G]}(\dot x)<\check\alpha_{\dot x}$. If $\dot y$ is any $Add(\omega, \omega_2)$-nice name for a real then there is an automorphism $\pi$ of $Add(\omega, \omega_2)$ that turns $\dot y$ into an $Add(\omega, \omega)$-name $\hat\pi(\dot y)$. But then $$1\Vdash f^{L_{\check\omega_2}(\mathbb R^\dot G)[\dot G]}(\hat\pi(\dot y))<\check\alpha_{\hat\pi(\dot y)}$$ implies $$ 1\Vdash f^{L_{\check\omega_2}(\mathbb R^{\widehat{\pi^{-1}}(\dot G)})[\widehat{\pi^{-1}}(\dot G)]}(\dot y)<\check\alpha_{\hat\pi(\dot y)}$$ Note that $f$ does not change as all the parameters in its definiton are in the ground model. Furthermore, it is easy to see that $$1\Vdash L_{\omega_2}(\mathbb R^{\widehat{\pi^{-1}}(\dot G)})[\widehat{\pi^{-1}}(\dot G)] = L_{\omega_2}(\mathbb R^{\dot G})[\dot G]$$ using that $\pi$ is definable in $M$. Thus $f^M(y)<\alpha_{\hat\pi(\dot y)}$ so that $f^M$ is bounded by $$\operatorname{sup}\{\alpha_{\dot x}\mid \dot x \text{ is a }Add(\omega, \omega)\text{-name}\}<\omega_2$$ We can use this to show that reflection is true in $M$.
Claim: For any $\in$-formula $\varphi$ there is a $M$-definable club of $\alpha<\omega_2$ with $\varphi$ absulute between $L_\alpha(\mathbb R^G)[G]$ and $M$.
Proof: We will do it by induction on the complexity of $\varphi$. The only difficult case is if $\varphi = \exists x \psi(x)$. Note that in $M$, any set is the surjective image of $\mathbb R^G\times\omega_1$ in $M$ (as all $L_\alpha(\mathbb R^G)[G]$, $\alpha<\omega_2$ are such an image). Even better, $\omega_1$ is the surjective image of $\mathbb R^G$ in $M$ so that all sets are the surjective image of a function with domain $\mathbb R^G$. Let $D$ be an $M$-definable club that witnesses our induction hypothesis for $\psi$. We can close some $L_\beta(\mathbb R^G)[G]$ with $\beta\in D$ under witnesses for $\varphi$ as follows: Find a surjection $f:\mathbb R^G\rightarrow [L_\beta(\mathbb R^G)[G]]^{<\omega}$ in $M$ and map a real $x$ to the least level of the hierachy in which a witness to $\varphi(\vec p)$ (if there is one) exists where $f(x)=\vec p$. This function must be bounded so that there is a least $\gamma\in D$ so that $L_\gamma(\mathbb R^G)[G]$ contains witnesses for $\varphi$ with parameters from $L_\beta(\mathbb R^G)[G]$. This map $\beta\mapsto \gamma$ is definable in $M$ and the closure points of this map form a club with the desired properties.$\Box$
Now replacement (and even better, collection) is an easy consequence of the reflection principle above.
Choice in M: The axiom of choice in $M$ (in terms of existence of choice functions) follows by somewhat similar arguments:
First note that for any $\alpha<\omega_2$ there is a map $C_\alpha:\mathbb R^G\rightarrow\mathbb R^G$ in $M$ so that $C_\alpha(x)$ is (better: codes canonically a) $Add(\omega, \omega)$-generic over $L[G\upharpoonright\alpha, x]$ for any $x$. For any $x\in\mathbb R^G$ there is some $\alpha\leq\beta<\alpha+\omega_1$ so that the section of $G$ starting at $\beta$ with length $\omega$ is generic over $L[G\upharpoonright\alpha][x]$. Let $C_\alpha(x)$ be this section with $\beta$ as small as possible. Then $C_\alpha$ is definable over $(L_{\alpha+\omega_1}(\mathbb R^G)[G];\in, G)$ and hence is in $M$.
[Here the use of the predicate $G$ is crucial! No such function $C$ exists in $L_{\omega_2}(\mathbb R^G)$, so in particular the map $x\mapsto \{y\mid y$ is $Add(\omega, \omega)$-generic over $L[x]\}$ does not have a choice function there.]
Now let $f\in M$ be a function we want to find a choice function for. Since replacement holds in $M$ and all sets there are surjective images of $\mathbb R^G$ we may assume that $f:\mathbb R\rightarrow\mathcal P(\mathbb R)^M\setminus\{\emptyset\}$. Let $\alpha<\omega_2$ be large enough so that $f$ is definable in $M$ from paraters in $L[G\upharpoonright\alpha]$ (remeber that these parameters can be chosen to be ordinals, reals and initial segments of $G$ so such an $\alpha$ exists).
Claim: If $x\in \mathbb R^G$ and $g\in M$ is $Add(\omega,\omega)$-generic over $L[G\upharpoonright\alpha, x]$ then $f(x)\cap\mathbb R^{L[G\upharpoonright\alpha, x, g]}\neq\emptyset$.
Proof: For notational reasons we assume $\alpha=0$, so that $L[G\upharpoonright\alpha] = L$. Work in $L[x]$. Note that $L[x]$ is either $L$ or a Cohen extension thereof. Also $g$ is generic over $L[x]$. Thus $G$ factors into $h\times H^\prime$ so that $L[x]=L[h]$ and $H^\prime$ is $Add(\omega, \omega_2)$-generic over $L[x]$ with $L[G]=L[x][H^\prime]$. Similarly, we can take an isomorphis $\mu:Add(\omega, \omega_2)\rightarrow Add(\omega, \omega)\times Add(\omega, \omega_2)$ so that $H^\prime$ factors again (via $\pi^{-1}$) into $g\times H$. In $L[x]$ we can find a nice name $\dot y$ so that $$1\Vdash^{L[x]}_{Add(\omega, \omega_2)} \dot y \in f(\check x)^{L_{\check\omega_2}(\mathbb R^{\dot G})[\dot G]}$$ Using the ccc, we can further find an automorphism $\pi$ of $Add(\omega, \omega_2)$ so that $\mu\circ\pi$ shifts $\dot y$ completely into the first factor $Add(\omega, \omega)$. With the same strategy as in the boundedness arguments, using that all parameters defining $f$ are in the ground model, we can see that in fact $$z=\widehat{\mu\circ\pi}(\dot y)^{g\times H^\prime}\in f(x)$$ But the evaluation of $\widehat{\mu\circ\pi}(\dot y)$ depends only on $g$ so that $z\in L[x][g]$ witnesses the claim to hold.
If $\alpha\neq 0$, replace $L$ by $L[G\upharpoonright\alpha]$ in the argument above, work from there and use that $L[G]$ is still a $Add(\omega,\omega_2)$-extension from there. $\Box$
It is now straight forward to define a choice function for $f$ in $M$: Just map a real $x$ to the $<_{L[x, C_\alpha(x)]}$-least element of $f(x)$ (where $<_{L[x, C_\alpha(X)]}$ denotes the canonical wellorder on the reals of $L[x, C_\alpha(x)]$).
(Edit: This replaces an earlier attempt I had in the opposite direction, which was bogus, as pointed out by @HarryWest, and to which some comments below refer.)
(Edit 2: It now gets weak compactness as a consistency strength lower bound.)
(Edit 3: A part initially missing at the end of the Subcase 2.2 argument has been filled in.)
(Edit 4: 2 observations specifically on $(\kappa,\kappa^{++})$-compactness added at the end.)
Remark: It looks like @HarryWest has already answered the original question. Below I work more directly with $\mathcal{R}$-satisfiability and extract some more strength. I don't know to what extent Harry's method would also lead to the following.
Remark 2: The reader unfamiliar with the Dodd-Jensen core model $K^{\mathrm{DJ}}$, but familiar with $0^\sharp$, should replace "there is an inner model with a measurable cardinal" in the claim below with "$0^\sharp$ exists", and replace $K^{\mathrm{DJ}}$ with $L$ throughout; this results in a weaker result, but the proof is essentially identical, and it suffices for Corollaries 1 and 3.
Claim. Assume ZFC. Let $\kappa$ be an uncountable cardinal such that $\mathcal{R}$-satisfiability is $(\kappa,\kappa^+)$-compact. Then there is a weakly inaccessible cardinal $\mu\leq\kappa$, and either:
there is an inner model with a measurable cardinal, or
letting $K=K^{\mathrm{DJ}}$ be the Dodd-Jensen core model below a measurable, $K\models$"$\kappa$ is weakly compact", and if $2^\gamma\leq\kappa$ for all $\gamma<\kappa$ then $\kappa$ is weakly compact.
Corollary 1. ZFC + "There is an uncountable cardinal $\kappa$ such that $\mathcal{R}$-satisfiability is $(\kappa,\kappa^+)$-compact" is equiconsistent with ZFC + "There is a weakly compact cardinal".
Corollary 2. Assume ZFC + GCH + "there is no inner model with a measurable cardinal". Let $\kappa$ be an uncountable cardinal. Then $\mathcal{R}$-satisfiability is $(\kappa,\kappa^+)$-compact iff $\kappa$ is weakly compact.
Corollary 3. Assume ZFC + GCH + "$0^\sharp$ does not exist". Let $\kappa$ be an uncountable cardinal. Then $\mathcal{R}$-satisfiability is $(\kappa,\kappa^+)$-compact iff $\kappa$ is weakly compact.
(Note that the converse direction of Corollaries 2 and 3, i.e. if $\kappa$ is weakly compact then $\mathcal{R}$-satisfiability is $(\kappa,\kappa^+)$-compact, holds in general.)
Proof of Claim: Fix $\kappa$. Except for the proof that there is a weakly inaccessible cardinal $\leq\kappa$, we may assume there is no inner model with a measurable cardinal, so write $K=K^{\mathrm{DJ}}$. (As mentioned in Remark 2, the reader unfamiliar with $K$ but familiar with $0^\#$ should just assume $0^\#$ does not exist, in which case $K=L$.) I will formally assume this, but it is easy to drop this assumption and still get the weakly inaccessible $\mu\leq\kappa$.
The plan is to find a reasonable elementary substructure $\bar{\mathcal{H}}$ of some fragment of $V$ and an elementary embedding $j:\bar{\mathcal{H}}\to M$ with a critical point $\mu$, and use this to get the desired conclusions.
If there is $\gamma<\kappa$ such that $2^\gamma\geq\kappa$ then letting $\gamma$ be least such, fix a sequence $\left<A_\alpha\right>_{\alpha<\kappa}$ of pairwise distinct subsets of $\gamma$. Otherwise let $A_\alpha=\emptyset$ for all $\alpha<\kappa$. Let $\mathcal{H}=(\mathcal{H}_{\kappa^+},\vec{A},<^*)$, where $<^*$ is a wellorder of $\mathcal{H}_{\kappa^+}$ (the set of all sets hereditarily of cardinality $\leq\kappa$).
We will build a theory $T$, of size $\kappa$, such that every sub-theory of size ${<\kappa}$ is $\mathcal{R}$-realizable. Basically, we want $T$ to describe a model which contains a version of $\mathcal{H}$ as an element, together with the statement "$\kappa$ is not a cardinal". The theory $T$ will use primary constants $\dot{\vec{\alpha}}$ for certain sequences $\vec{\alpha}$ of ordinals ${<\kappa^+}$. The sequences $\vec{\alpha}$ used we call the relevant sequences, and which are considered relevant depends on the following cases. If (i) $\gamma^{\omega}\leq\kappa$ for all $\gamma<\kappa$ then there are only $\kappa$-many $\omega$-sequences $\vec{\alpha}\in{^\omega}\kappa$ such that $\vec{\alpha}$ is bounded in $\kappa$ (and of course if $\mathrm{cof}(\kappa)>\omega$, "bounded in $\kappa$" can be struck out), and in this case these sequences $\vec{\alpha}$ are the relevant ones. Suppose instead (ii) there is $\gamma<\kappa$ such that $\gamma^{\omega}>\kappa$. If (ii.1) $\kappa^{+K}<\kappa^+$ then the relevant sequences are just the finite tuples $\vec{\alpha}\in\kappa^{<\omega}$. Suppose instead (ii.2) $\kappa^{+K}=\kappa^+$. In this case we need to be a little more careful. Fix from now on an ordinal $\eta\in(\kappa,\kappa^+)$ of cofinality $\kappa$ and such that letting $\bar{\mathcal{H}}=\mathrm{Hull}^{\mathcal{H}}(\eta)$, we have $\eta=\bar{\mathcal{H}}\cap\eta$. Then the relevant sequences are the finite tuples $\vec{\alpha}\in\eta^{<\omega}$. (Also for the proof that there is a weakly inaccessible cardinal $\mu\leq\kappa$, i.e. without assuming $K$ exists, in case (ii), proceed as in (ii.1).)
Using the process in the answer here: https://mathoverflow.net/questions/394526/is-this-compactness-property-for-satisfiability-on-mathbbr-consistent, we then also augment the theory with some secondary constants, so as to allow us to talk about subsets of $\omega$ coded by reals, and to make arithmetic statements about those coded sets using (infinitely many) statements in $T$. We will skip the details of those extra constants, and just directly make arithmetic statements about the coded subsets of $\omega$.
For each relevant $\vec{\alpha}$, let $t_{\vec{\alpha}}$ be the full theory of the single parameter $\vec{\alpha}$ in the structure $\mathcal{H}$ (note this includes predicates for $\vec{A}$ and $<^*$). Add the following statements to $T$ (they mostly refer to theories coded by reals):
$\dot{\vec{\alpha}}$ codes a consistent complete theory $u_{\dot{\vec{\alpha}}}$ in the language of set theory augmented with symbols $\hat{\vec{\beta}},\hat{\mathcal{H}},\hat{f},\hat{\kappa},\hat{\xi}$,
$u_{\dot{\vec{\alpha}}}$ contains the formula "$\hat{\mathcal{H}}=(\mathcal{J},<',\vec{A}')$ is a structure with transitive universe $\hat{\mathcal{J}}$, $<'$ is a wellorder of $\mathcal{J}$, and $\hat{\kappa}$ is the largest cardinal of $\mathcal{J}$",
$u_{\dot{\vec{\alpha}}}$ contains the formula "$V=L(\hat{\mathcal{H}},\hat{f})$ and $\hat{\xi}<\hat{\kappa}$ and $\hat{f}:\hat{\xi}\to\hat{\kappa}$ is a surjection",
the model determined by $u_{\dot{\vec{\alpha}}}$ has standard $\omega$,
and for each formula $\varphi[\vec{\alpha}]\in t_{\vec{\alpha}}$, add the statement
- $u_{\dot{\vec{\alpha}}}$ contains the formula "$\hat{\mathcal{H}}\models\varphi[\dot{\vec{\beta}}]$"
to $T$. Moreover, if $\vec{\gamma}$ is also relevant and $\mathrm{rg}(\vec{\alpha})\subseteq\mathrm{rg}(\vec{\gamma})$ and in the $\omega$-sequence case, $\vec{\alpha}$ itself is easily computed from $\vec{\gamma}$ (say there is a recursive function $i:\omega\to\omega$ such that $\vec{\alpha}_n=\vec{\gamma}_{i(n)}$), then we add the formula
- $u_{\dot{\vec{\alpha}}}$ is the theory induced by $u_{\dot{\vec{\gamma}}}$ (according to how $\vec{\alpha}$ is computed from $\vec{\gamma}$)
to $T$.
Now let $S\subseteq T$ be a sub-theory of size ${<\kappa}$. We find an $\mathcal{R}$-realization of $S$. Let $C$ be the set of relevant sequences $\vec{\alpha}$ used in $S$; so $C$ has size ${<\kappa}$, and $C\subseteq\mathcal{H}$. Let $H=\mathrm{Hull}^{\mathcal{H}}(C)$; that is, the structure whose universe is the set of all elements of $\mathcal{H}_{\kappa^+}$ definable over $\mathcal{H}$ from parameters in $C$ (using the predicates of $\mathcal{H}$), and with predicates being the restrictions of those of $\mathcal{H}$. So $H\preccurlyeq\mathcal{H}$. Let $\bar{H}$ be the transitive collapse of $H$. Let $\pi:\bar{H}\to H$ be the uncollapse map. Let $\bar{\kappa}=\pi^{-1}(\kappa)$ etc. By enlarging $S,C$ if necessary, we may assume that $\bar{\kappa}$ has cardinality ${\xi<\bar{\kappa}}$. So let $f:\xi\to\bar{\kappa}$ be a surjection. Let $N=L_\beta(\bar{H},f)$ for some ordinal $\beta>0$.
Now for each $\vec{\alpha}\in C$, let $u_{\vec{\alpha}}$ be the theory in $N$ of the parameters $\bar{\vec{\alpha}},\bar{H},f,\xi,\bar{\kappa}$, where bars denote preimage under $\pi$. Then note that by interpreting $\dot{\vec{\alpha}}$ as the real naturally coding $u_{\vec{\alpha}}$, we get an $\mathcal{R}$-realization of $S$.
By $(\kappa,\kappa^+)$-compactness, we can fix an $\mathcal{R}$-realization $\mathcal{R}^+$ of $T$. Let $u_{\vec{\alpha}}$ be the theory coded by $\dot{\vec{\alpha}}^{\mathcal{R}^+}$. Note that we can define a term model $M$, pointwise definable from constants $\widetilde{\vec{\alpha}}$ (for relevant sequences $\vec{\alpha}$ as before) and constants $\widetilde{\mathcal{H}},\widetilde{f},\widetilde{\xi},\widetilde{\kappa}$, and such that $u_{\vec{\alpha}}$ is just the theory in $M$ of $\widetilde{\vec{\alpha}}^M,\widetilde{\mathcal{H}}^M,\widetilde{f}^M,\widetilde{\xi}^M,\widetilde{\kappa}^M$ (the super-$M$ denotes the interpretation of a constant in $M$).
Let $\bar{\mathcal{H}}=\mathrm{Hull}^{\mathcal{H}}(C)$ where $C$ is the set of all relevant sequences (the hull is uncollapsed); note that $\bar{\mathcal{H}}$ is in fact transitive (so actually it's the same as the collapsed hull) so $\bar{\mathcal{H}}\preccurlyeq\mathcal{H}$ (here $\bar{\mathcal{H}}$ is a structure with predicates induced by those of $\mathcal{H}$) (and note that in case (ii.2), what we defined as $\bar{\mathcal{H}}$ earlier is the same as the model we have just now defined, and so in this case $\mathrm{OR}\cap\bar{\mathcal{H}}=\eta$). We have $\kappa+1\subseteq\bar{\mathcal{H}}$. (But $\kappa^+\not\subseteq\bar{\mathcal{H}}$, since $\bar{\mathcal{H}}$ has cardinality $\kappa$.) Note that there is an elementary embedding $j:\bar{\mathcal{H}}\to\widetilde{\mathcal{H}}^M$ given by setting $j(\vec{\alpha})=\widetilde{\vec{\alpha}}^M$ for each relevant $\vec{\alpha}$.
We have $j(\kappa)=\kappa^M$. Since $\kappa$ is a cardinal in $V$, but $\kappa^M$ is not a cardinal in $M$ (though it is of course a cardinal in $\widetilde{\mathcal{H}}^M$), $j$ has a critical point $\mu\leq\kappa$ (here $M$ might be illfounded). So $\mu$ is a regular cardinal in $\bar{\mathcal{H}}$, and hence also regular in $V$. Note that $\mu>\omega$, as $M$ has standard $\omega$, as this held for each sub-model coded by $u_{\vec{\alpha}}$. Note that $\mu$ is not a successor cardinal (if $\mu=\gamma^+$ then $$\gamma^+=\mu<j(\mu)=j(\gamma^+)=j(\gamma)^{+M}=\gamma^{+M},$$ so $\mu$ is collapsed in $M$, a contradiction). So $\mu$ is weakly inaccessible, in particular giving the existence of a weakly inaccessible $\leq\kappa$ (we didn't yet use $K$). So from now on we do assume we have $K$, i.e. there is no inner model with a measurable.
Case 1: $\gamma^{\omega}\leq\kappa$ for all $\gamma<\kappa$.
So the relvant sequences are the bounded-in-$\kappa$ $\omega$-sequences $\vec{\alpha}$. We consider two subcases.
Subcase 1.1: $\mu<\kappa$.
In this case there is an inner model with a measurable cardinal. For if not, then the Dodd-Jensen core model $K$ (below an inner model with a measurable cardinal) exists. So there is no transitive proper class $K'$ and non-trivial elementary $k:K\to K'$. And $K\models GCH$, and $\mathcal{P}(\mu)\cap K\subseteq\bar{\mathcal{H}}$.
Let $\delta$ be some "ordinal" of $M$ such that $\beta<\delta<j(\mu)$ for all $\beta<\mu$ (it doesn't matter whether $M$ is wellfounded; but we will use that $M$ has wellfounded $\omega$). Let $U$ be the $K$-ultrafilter derived from $j$ with seed $\delta$. Then $\mathrm{Ult}(K,U)$ is wellfounded. For suppose not, and let $\left<f_n,X_n\right>_{n<\omega}\subseteq K$ be such that $f_n:\mu\to\mathrm{OR}$ and $X_n\subseteq\mu$ and $X_n\in U$ and $f_{n+1}(\alpha)<f_n(\alpha)$ for all $\alpha\in X_n$, for all $n<\omega$. Let $\vec{X}=\left<X_n\right>_{n<\omega}$. Because $K|\mu^{+K}$ is definable over $\mathcal{H}_\kappa$ and satisfies "$V=\mathrm{HOD}$", each $X_n$ is specified by some ordinal $\alpha<\mu^+$, so there is a relevant $\vec{\alpha}$ such that $\left<X_n\right>_{n<\omega}$ is defined from $\vec{\alpha}$. Let $X=\bigcap_{n<\omega}X_n$. So $X\in\bar{\mathcal{H}}$, and (as $M$ has wellfounded $\omega$) $X\in U$, and in particular $X\neq\emptyset$. But then letting $\beta\in X$, we have $f_{n+1}(\beta)<f_n(\beta)$ for all $n<\omega$, a contradiction.
So $K'=\mathrm{Ult}(K,U)$ is wellfounded, but then the ultrapower map $k:K\to K'$ is non-trivial, a contradiction.
Subcase 1.2: $\mu=\kappa$.
So all $\omega$-sequences $\subseteq\kappa$ are relevant, and it easily follows that $M$ is closed under $\omega$-sequences, and hence wellfounded. We claim that $K\models$"$\kappa$ is weakly compact". For since $\kappa$ is weakly inaccessible, it is inaccessible in $K$. Suppose $\kappa$ is not weakly compact in $K$. We have $K_\kappa=V_\kappa^K\subseteq\bar{\mathcal{H}}$ and $V_{\kappa+1}^K\subseteq\mathcal{H}$, and so $\bar{\mathcal{H}}\models$"$\kappa$ is not weakly compact in $K$". Let $A\in V_{\kappa+1}^K$ with $A\in\bar{\mathcal{H}}$ and $\varphi$ be a $\Pi_1$ formula such that $(V_{\kappa+1}^K,A)\models\varphi$ but there is no $\kappa'<\kappa$ such that $(V_{\kappa'+1}^K,A\cap V_{\kappa'}^K)\models\varphi$. So the same holds for $j(\kappa)$ in $M$. Since $M$ is wellfounded, we have $\kappa\in M$, and can apply the statement at $\kappa'=\kappa$ in $M$, which implies $(V_{\kappa+1}^{K^M},j(A)\cap V_{\kappa}^{K^M})\models\neg\varphi$, and so in fact $(V_{\kappa+1}^{K^M},A)\models\neg\varphi$. But $M$ is closed under $\omega$-sequences, which implies that $K^M$ is iterable, and so $K^M|\kappa^{+K^M}$ is a segment of $K$ so $V_{\kappa+1}^{K^M}\subseteq K^M$, and since $\neg\varphi$ is $\Sigma_1$, therefore $(V_{\kappa+1}^K,A)\models\neg\varphi$, a contradiction.
Now suppose $2^\gamma\leq\kappa$ for all $\gamma<\kappa$. We first claim that $\kappa$ is actually inaccessible. For suppose not, and let $\gamma<\kappa$ be least such that $2^\gamma\geq\kappa$; by our assumption then, $2^\gamma=\kappa$. Thus, we chose $\vec{A}$ to enumerate all subsets of $\gamma$, in a one-to-one fashion. But then $j(\vec{A})$ properly extends $\vec{A}$ with new subsets of $\gamma$, a contradiction. Now one can show that $\kappa$ is weakly compact via an argument just like that used for $K$ above, but now in $V$ (and it is easier).
Case 2: Otherwise (there is $\gamma<\kappa$ such that $\gamma^{\omega}>\kappa$).
This case will be dealt with similarly to the previous one, but it is a little subtler. The relevant sequences are the finite tuples $\vec{\alpha}\in\eta^{<\omega}$, where $\eta=\kappa$ in case (ii.1).
Subcase 2.1: $\mu<\kappa$.
Define $\delta,U$ as in Subcase 1.1. We claim again that $\mathrm{Ult}(K,U)$ is wellfounded, which is again enough. Suppose not and let $\left<f_n,X_n\right>_{n<\omega}\subseteq K$ be as before. We have $X_n\subseteq\mu$. Let $\alpha_n$ be the rank of $X_n$ in the $K$-constructibility order. Then $\alpha_n<\mu^{+K}\leq\mu^+\leq\kappa$. By covering for $K$, there is a set $\mathcal{X}\in K$, of cardinality $\leq\aleph_1^V$, such that $X_n\in\mathcal{X}$ for each $n$, and $Y\subseteq\mu$ for each $Y\in\mathcal{X}$. Since $\aleph_2^V<\mu$, the usual argument shows that $Z=\{Y\in\mathcal{X}\bigm|Y\in U\}$ is in $K$. (That is, consider $\{Y\in j(\mathcal{X})\bigm|\delta\in Y\}\in M$, and use the agreement between $M$ and $\bar{\mathcal{H}}$ below $\mu$.) But then also since $\aleph_2^V<\mu$, it follows that $\bigcap Z\neq\emptyset$. Therefore $\bigcap_{n<\omega}X_n\neq\emptyset$, so now we can argue for contradiction like in Subcase 2.1.
Subcase 2.2: $\mu=\kappa$.
In this case we show that $\kappa$ is weakly compact in $K$.
Suppose (ii.2) holds, so $\kappa^{+K}=\kappa^+$ and $\bar{\mathcal{H}}=\mathrm{Hull}^{\mathcal{H}}(\eta)$, and $\mathrm{cof}(\eta)=\kappa$. Suppose $\kappa$ is not weakly compact in $K$. Then $K|\eta=K^{\bar{\mathcal{H}}}\models$"$\kappa$ is not weakly compact". So fix a counterexample $A,\varphi\in K|\eta$.
Define $U,\delta$ as before. We claim that $\mathrm{Ult}(K|\eta,U)$ is wellfounded. So suppose otherwise and let $\left<f_n,X_n\right>_{n<\omega}$ be a counterexample $\subseteq K|\eta$. Let $\alpha_n$ be the rank of $X_n$ in the order of constructibility of $K$. Let $\mathcal{X}=\{\alpha_n\bigm|n<\omega\}$. So by covering, there is $\mathcal{Y}\subset\mathrm{OR}$ such that $\mathcal{Y}\in K$ and $\mathcal{X}\subseteq\mathcal{Y}$ and $\mathcal{Y}$ has cardinality $\leq\aleph_1^V$. It suffices to see there is such a $\mathcal{Y}\in K|\eta$, as then we can argue as before. Since $\kappa$ is the largest cardinal of $K|\eta$, there are cofinally many $\beta<\eta$ such that $K|\beta$ projects to $\kappa$, which means here that there is a surjection $f:\kappa\to K|\beta$ which is definable without parameters over $K|\beta$. Fix such a $\beta,f$ with $X_n\in K|\beta$ for all $n$; this exists because $\mathrm{cof}(\eta)=\kappa$ and $\kappa$ is weakly inaccessible. So $f\in K|(\beta+1)\subseteq K|\eta$. Let $\mathcal{X}'$ be the set of all $\alpha<\kappa$ such that $f(\alpha)\in\mathcal{X}'$ and for no $\alpha'<\alpha$ is $f(\alpha')=f(\alpha)$. So $\mathcal{X}'\subset\kappa$ and $\mathcal{X}'$ is countable. So by covering, there is $\mathcal{Y}'\in K$ with $|\mathcal{Y}'|\leq\aleph_1^V$ and $\mathcal{X}'\subseteq\mathcal{Y}'$. Since $\kappa$ is weakly inaccessible, $\mathcal{Y}'\in K|\kappa$. But then $\mathcal{Y}=f``\mathcal{Y}'$ covers $\mathcal{X}$ and $\mathcal{Y}\in K|(\beta+1)\subseteq K|\eta$, as desired.
Also, $N=\mathrm{Ult}(K|\eta,U)$ is iterable. For it suffices to see that all countable elementary substructures of $N$ are iterable. For this, given $\left<f_n\right>_{n<\omega}\subseteq K$, it suffices to see that $\bar{N}=\mathrm{Hull}^N(\{[f_n]\bigm|n<\omega\})$ is iterable (where $f_n$ represents $[f_n]\in N$). But we can find sets $X_n\in K|\eta$ such that the $\Sigma_n$-elementary theory of $(f_0(\alpha),\ldots,f_n(\alpha))$ in $K|\eta$ is independent of $\alpha\in X_n$, and then arguing as before, $\bigcap_{n<\omega}X_n\neq\emptyset$, so letting $\alpha\in\bigcap_{n<\omega}X_n$, $\bar{N}$ is isomorphic to $\mathrm{Hull}^{K|\eta}(\{f_n(\alpha)\bigm|n<\omega\})$, and is therefore iterable.
By the iterability, $N|\kappa^{+N}=K|\kappa^{+N}$, and so the failure of weak compactness now leads to a contradiction like before (applying the ultrapower map $i^{K_\eta}_U:K|\eta\to N$).
Finally if (ii.1) holds, it is almost the same as for (ii.2), but easier: note that $\kappa^{+K}<\eta=\mathrm{OR}\cap\bar{\mathcal{H}}$ in this case, so taking any (small) covering set $\mathcal{Y}\in K$ for a countable set $\mathcal{X}$, we get $\mathcal{Y}\in\bar{\mathcal{H}}$, which is again enough.
Regarding specifically $(\kappa,\kappa^{++})$-compactness of $\mathcal{R}$-satisfiability:
Observation 1: Assuming ZFC + $\kappa$ is uncountable + $\mathcal{R}$-satisfiability is $(\kappa,\kappa^{++})$-compact, we get an inner model with a measurable, by arguing much as above, but allowing relevant sequences to take values anywhere ${<\kappa^+}$; this leads to a model $\bar{\mathcal{H}}\preccurlyeq\mathcal{H}$ with $\kappa^+\subseteq\bar{\mathcal{H}}$ and an embedding $j:\bar{\mathcal{H}}\to\widetilde{\mathcal{H}}^M$ at the end, and $j(\kappa)>\kappa$. This then gives an inner model with a measurable like when $\mu<\kappa$ in the preceding proof.
Observation 2: In the converse direction: Assume ZFC + GCH + $\kappa$ is $\kappa^+$-supercompact. Let $G$ be generic over $V$ for adding a $\kappa^{++}$-seqence of Cohen reals with the finite support product. Then $V[G]\models$"$2^{\aleph_0}=\kappa^{++}$ and $\mathcal{R}$-satisfiability is $(\kappa,\kappa^{++})$-compact".
This can be shown using a small variant of the argument in Update 2 of my answer to https://mathoverflow.net/questions/394526/is-this-compactness-property-for-satisfiability-on-mathbbr-consistent, which does the analogous thing for $(\kappa,\kappa^+)$-compactness from weak compactness.
Best Answer
Since $\omega_2$ is regular, any function from $\omega$ to $\omega_2$ is bounded. Thus $\omega_2^\omega=\bigcup_{\alpha<\omega_2}\alpha^\omega$. The cardinality of this is just $\omega_2\cdot\sup_{\alpha<\omega_2}|\alpha^{\omega}|$. Since $|\alpha^{\omega}|\le|\omega_1^{\omega}|=\omega_1$ under $\mathsf{CH}$ for $\alpha<\omega_2$, $\omega_2^\omega=\omega_2\cdot\omega_1=\omega_2$.