It should be noted that already in $\mathbf{Set}$, the free functor $\mathbf Z^{(-)} \colon \mathbf{Set} \to \mathbf{Ab}$ does not preserve cofiltered limits. For a cofiltered diagram $D \colon \mathcal I \to \mathbf{Set}$, write $S_i$ for its value at $i \in \mathcal I$, write $S$ for its limit, and write $\pi_i \colon S \to S_i$ for the canonical projection.
There is always a map
\begin{align*}
\phi \colon \mathbf Z^{(S)} &\to \lim_\leftarrow \mathbf Z^{(S_i)}\\
\sum_{k=1}^n n_k s_k &\mapsto \left( \sum_{k=1}^n n_k\pi_i(s_k) \right)_i.
\end{align*}
Denote its $i^{\operatorname{th}}$ component by $\phi_i$. For a set $X$ and an element $z=\sum_{k=1}^n n_k x_k \in \mathbf Z^{(X)}$ with $x_k \neq x_{k'}$ for $k \neq k'$, write $\operatorname{Supp}(z)$ for $\{x_1,\ldots,x_n\}$, and denote by $|z|$ its cardinality. If $f \colon X \to Y$ is a map, we denote the induced map $\mathbf Z^{(X)} \to \mathbf Z^{(Y)}$ by $f$ as well (by abuse of notation). For $z \in \mathbf Z^{(X)}$, we have $\operatorname{Supp}(f(z)) \subseteq f(\operatorname{Supp}(z))$, so $|f(z)| \leq |z|$ with equality if and only if $f_* \colon \operatorname{Supp}(z) \to \operatorname{Supp}(f(z))$ is a bijection.
Lemma. The map $\phi$ is injective, and its image consists of those $(x_i)_{i \in \mathcal I}$ for which there exists $n \in \mathbf Z$ with $|x_i| \leq n$ for all $i \in \mathcal I$.
In other words, the image consists of the sequences $(x_i)_i$ of bounded support.
Proof. For injectivity, if $x = \sum_{k=1}^n n_ks_k \in \ker(\phi)$ is such that $s_k \neq s_{k'}$ for $k \neq k'$, then there exists $i \in \mathcal I$ such that $\pi_i(s_k) \neq \pi_i(s_{k'})$ for $k \neq k'$ (here we use that the sum is finite and that $\mathcal I$ is cofiltered). Then $\phi_i(x) = \sum_{k=1}^n n_k\pi_i(s_k)$ is zero by assumption, so all $n_k$ are zero.
For the image, it is clear that $|\phi_i(x)| \leq n$ for all $i \in \mathcal I$ if $x \in \mathbf Z^{(S)}$ has $|x| = n$. Conversely, if $|x_i| \leq n$ for all $i \in \mathcal I$, then decreasing $n$ if necessary, we may assume $|x_{i_0}| = n$ for some $i_0 \in \mathcal I$. Then $|x_i| = n$ for all $f \colon i \to i_0$ since $n = |x_{i_0}| = |D(f)(x_i)| \leq |x_i| \leq n$. Replacing $\mathcal I$ by the coinitial segment $\mathcal I/i_0$ we may therefore assume $|x_i| = n$ for all $i \in \mathcal I$.
Constancy of $|x_i|$ means that for every morphism $f \colon i \to j$ in $\mathcal I$, the map $f_* \colon \operatorname{Supp}(x_i) \to \operatorname{Supp}(x_j)$ is a bijection. Setting $T = \lim\limits_\leftarrow \operatorname{Supp}(x_i)$, we see that each projection $\pi_i \colon T \to \operatorname{Supp}(x_i)$ is a bijection. Functoriality of the limit gives an injection $T \hookrightarrow S$, giving elements $s_1,\ldots,s_n \in S$ such that $\pi_i(\{s_1,\ldots,s_n\}) = \operatorname{Supp}(x_i)$ for all $i \in \mathcal I$. The coefficients must also be constant under the bijections $\operatorname{Supp}(x_i) \to \operatorname{Supp}(x_j)$, so we get an element $x = \sum_{k=1}^n n_ks_k \in \mathbf Z^{(S)}$ with $\phi(x) = (x_i)_i$. $\square$
Example. An example where $\phi$ is not surjective: let $S = \mathbf Z_3$ with $S_i = \mathbf Z/3^i$. Define the element $(x_i)_i \in \prod_i \mathbf Z^{(S_i)}$ where $x_i \in \mathbf Z^{(S_i)} = \mathbf Z^{S_i}$ has coordinates (for $k \in \{0,\ldots,3^i-1\}$) given by
$$x_{i,k} = \begin{cases} 1, & \text{the first $3$-adic digit of } k \text{ is } 1, \\ -1, & \text{the first $3$-adic digit of } k \text{ is } 2, \\ 0, & k=0.\end{cases}$$
These form an element of $\lim\limits_\leftarrow \mathbf Z^{(S_i)}$: any fibre of $\mathbf Z/3^{i+1} \to \mathbf Z/3^i$ above $k \in \{0,\ldots,3^i-1\}$ consists of $\{k,3^i+k,2 \cdot 3^i+k\}$, of which $x_{i+1,k} = x_{i,k}$ and the others are $1$ and $-1$ since $3^i+k$ starts on $1$ and $2 \cdot 3^i+k$ starts on $2$.
Since $\operatorname{Supp}(x_i) = S_i \setminus \{0\}$, we see that $(x_i)_i$ does not have bounded support, hence is not in the image of $\phi$.
Corollary. For any presheaf topos $\mathbf T = \mathbf{PSh}(\mathscr C) = [\mathscr C^{\operatorname{op}},\mathbf{Set}]$ on a small nonempty category $\mathscr C$, the free functor $\mathbf Z^{(-)} \colon \mathbf T \to \mathbf{Ab}(\mathbf T)$ does not preserve cofiltered limits.
Proof. Let $f \colon \mathscr C \to *$ be the map to the terminal category $*$, on which $\mathbf{PSh}(*) = \mathbf{Set}$. Note that $f$ has a section $g$ since $\mathscr C$ is nonempty. We saw above that in $\mathbf{Set}$, the free functor $\mathbf Z^{(-)}$ does not preserve cofiltered limits. The pullback $f^* \colon \mathbf{Set} \to \mathbf{PSh}(\mathscr C)$ takes a set $S$ to the constant sheaf $\underline{S}$ on $\mathscr C$. Since limits, colimits, and free abelian group objects in presheaf categories are pointwise, $f^*$ commutes with formation of limits, colimits, and free abelian group objects. Thus pulling back everything along $f^*$ gives the natural map $\underline{\mathbf Z}^{(\underline S)} \to \lim\limits_\leftarrow \underline{\mathbf Z}^{(\underline S_i)}$, which is not an isomorphism since it isn't after applying the section $g^*$ (this is just evaluation at an object $g(*) \in \mathscr C$). $\square$
The class of topoi where the free functor does not commute with cofiltered limits is probably much larger still. For instance, my example does not include condensed sets, which is close to a presheaf topos but not quite. I'm not even sure what a topos would look like where these do commute!
As Wojowu noted in the comments, one should really look at $T_1$ topological spaces. Consider the functors
\begin{align*}
G\!: \mathbf{Top}_{T_1} &\leftrightarrows \mathbf{Cond}_\kappa:\!F\\
X &\mapsto \big(\underline X \colon S \mapsto \operatorname{Cont}(S,X)\big)\\
T(*) &\leftarrow\!\shortmid T,
\end{align*}
where $T(*)$ is topologised with the quotient topology via the surjection
$$\pi \colon \coprod_{(S,f \in T(S))} S \to T(*)\tag{1}\label{1}$$
given on the component $f \in T(S)$ by $f \colon S \to T(*)$. By the Yoneda lemma, this really means that $f \colon h_S \to T$ is a morphism from the representable sheaf $h_S$ to $T$, and the map $S \to T(*)$ is the set-theoretic map $f_{\{*\}} \colon h_S(*) \to T(*)$. But there is the consistent abuse of notation to denote $h_S = \underline S$ as $S$. To see that (\ref{1}) is surjective, use $S = \{*\}$.
As Dylan Wilson noted in the comments, the unit $\eta \colon 1 \to GF$ is given by the natural transformation
\begin{align*}
(\eta_T)_S \colon T(S) &\to \operatorname{Cont}(S,T(*)) \\
f &\mapsto \pi \circ \iota_f,
\end{align*}
where $\iota_f \colon S \to \displaystyle\coprod_{(S',f')} S'$ is the insertion of the coordinate corresponding to $(S,f)$. This gives the maps
\begin{align*}
\phi\!: \operatorname{Hom}_{\mathbf{Cond}}(T,\underline X) &\leftrightarrows \operatorname{Cont}(T(*),X) :\!\psi \\
f &\mapsto f_{\{*\}} \\
g \circ \eta_T &\leftarrow\!\shortmid g.
\end{align*}
It remains to check that these are inverses. If $g \colon T(*) \to X$ is continuous, then $\phi(\psi(g)) \colon T(*) \to X$ is given by $\psi(g)_{\{*\}}$, i.e. the map taking $t \in T(*)$ to $g \circ \pi \circ \iota_f \colon \{*\} \to X$. But $\pi \circ g \colon \{*\} \to T(*)$ is just (the constant map with value) the point $t$ (this is how we checked surjectivity of $\pi$ earlier!), so $\psi(g)_{\{*\}}$ takes $t$ to $g(t)$, i.e. $\phi(\psi(g)) = g$.
Conversely, given a natural transformation $f \colon T \to \underline X$, we get another natural transformation $\psi(\phi(f)) \colon T \to \underline X$. Write $g \colon T(*) \to X$ for $\phi(f) = f_{\{*\}}$. If $S$ is extremally disconnected and $h \in T(S)$, then $\psi(g)_S$ takes $h$ to the composition
$$S \overset{\iota_h}\to \coprod_{(S',h')} S' \overset\pi\to T(*) \overset g\to X.$$
By definition, the composition $\pi \circ \iota_h \colon S \to T(*)$ is the map $h_{\{*\}} \colon h_S(*) \to T(*)$. Thus $\psi(g)_S(h)$ is the composition
$$h_S(*) \overset{h_{\{*\}}}\longrightarrow T(*) \overset{f_{\{*\}}}\longrightarrow X.$$
This is the same thing as $(fh)_{\{*\}} \colon h_S(*) \to \underline X(*)$, which is the continuous map $S \to X$ given by $f_S(h) \in \underline X(S)$. We conclude that $\psi(\phi(f))_S(h) = f_S(h)$, and since $S$ and $h$ are arbitrary that $\psi(\phi(f)) = f$. $\square$
Remark. Morally what's going on here is the following: since $T(*)$ has the quotient topology, a map $g \colon T(*) \to X$ is continuous if and only if each of the compositions $gf \colon S \to X$ are continuous with $S$ extremally disconnected and $f \colon S \to T$ an $S$-point of $T$. Thus, a continuous map $T(*) \to X$ is the same as continuous maps $S \to X$ for every $S \to T$, which is roughly what a natural transformation $T \to \underline X$ is.
Best Answer
The answer is no (in general) for question 0 and no for question 2. In the edit of the question I have explained why $S \mapsto X(S)/G$ is not a condensed set in general.
In the following sense, the right generalization of the orbit space seems to be the categorical quotient (thanks to Echo's comment, I got to know this definition), and the categorical quotient of an action of $G$ on $X$ is the sheafification of $Y:S \mapsto X(S)/G$.
To see this, let's start by denoting $Y^+$ as the sheafification of $Y$. Let $\pi: X \rightarrow Y$ and $i: Y \rightarrow Y^+$ be the projection and the canonical morphism, respectively. For every $g \in G$, it is immediate that \begin{align*} i \circ \pi \circ g = i \circ \pi \end{align*} Now, let $Z$ be a condensed set with a map $q:X \rightarrow Z$, such that for each $g \in G$, we have \begin{align*} q \circ g = q \end{align*} Then, for each profinite set $S$, we have $q_S \circ g = q_S$. We define \begin{align*} \phi_S: Y(S) = X(S)/G &\longrightarrow Z(S) \\ \overline{x} &\longmapsto q_S (x) \end{align*} It is easy to check that $\phi_S$ is well-defined and that it defines a natural transformation $\phi:Y \rightarrow Z$. Also, note that $\phi$ is the only option to commute this diagram. Therefore, by the universal property of the sheafification, there is a unique map $Y^+ \rightarrow Z$ commuting the desired diagram.