I think that is correct. One can also do that by considering orientation forms as follows.
Let $(M,\mathcal{O})$ be the connected smooth oriented manifold. By hypothesis, there exists a nonvanishing $n$-form $\omega$ on $M$ such that $\omega$ is positively oriented at each point. Any other choice of orientation $\widetilde{\mathcal{O}}$ for $M$ will induce a nonvanishing $n$-form $\widetilde{\omega}$ on $M$. Since $\omega = f \widetilde{\omega}$ for some smooth function $f : M \to \mathbb{R}$ such that $f(p) \neq 0$ for all $p \in M$, then by the connectedness of $M$, the image $f(M) \subseteq \mathbb{R}$ is a connected subset which does not contain $0$. I.e., the function $f$ either always positive or negative on $M$. If $f$ is positive, then $\omega_p$ and $\widetilde{\omega}_p$ determine the same orientation at $T_pM$ for each $p \in M$. Hence $\widetilde{\mathcal{O}} = \mathcal{O}$. So lets assume that $f$ is negative. By the similar argument $\omega_p$ and $\widetilde{\omega}_p$ determine different orientation at $T_pM$ for each $p \in M$. So $\mathcal{O}$ is different than $\widetilde{\mathcal{O}}$. By similar argument as above, any other orientation $\mathcal{O}'$ for $M$ will induce a non-vanishing $n$-form $\omega'$ such that $\omega' = g \omega$ for either $g : M \to \mathbb{R}$ is positive or negative function. If $g$ is positive, then $\mathcal{O}' = \mathcal{O}$. If $g$ is negative then $\mathcal{O}'$ different than $\mathcal{O}$ and then the product $gf$ is a positive function. Hence above relation $\omega' =g \omega= gf \widetilde{\omega}$ implies that $\mathcal{O} = \widetilde{\mathcal{O}}$. This proves that there are exactly two orientation on $M$.
Suppose $\mathcal{O}_1$ and $\mathcal{O}_2$ are orientation for $M$ and $\omega_1$ and $\omega_2$ are the orientation forms for $\mathcal{O}_1$ and $\mathcal{O}_2$ respectively. Let $\omega_1 = f \omega_2$. If they agree at a point $p \in M$, then the orientation forms is positive multiples of each other at $p$. This implies that $f$ is a positive function. Hence $\omega_1$ and $\omega_2$ determines the same orientation for each point in $M$, i.e., $\mathcal{O}_1 = \mathcal{O}_2$.
Palais' disc theorem is not relevant to the question of orientability. It is relevant in arguing that the (oriented) connected sum is independent of the chosen discs up to (oriented) diffeomorphism.
Let the set-up be as in your question. You can find open collar neighborhoods of $\varphi(S^{n-1})$ in $M\setminus\varphi(\mathrm{int} B^n)$ and $\varphi^{\prime}(S^{n-1})$ in $M^{\prime}\setminus\varphi^{\prime}(\mathrm{int} B^n)$, which glue together to give an open bicollar $U$ around the embedded copy of $S^{n-1}$ in $M\# M^{\prime}$. Now $M\#M^{\prime}$ is the union of the three open subsets $M\setminus\varphi(\mathrm{int} B^n)$, $U$ and $M^{\prime}\setminus\varphi^{\prime}(\mathrm{int} B^n)$ , each intersecting only the ones written adjacent to it. Assume you're given an orientation of $M$, hence an orientation of $M\setminus\varphi(\mathrm{int} B^n)$. The intersection $M\setminus\varphi(\mathrm{int} B^n)\cap U$ is connected (here, $n\ge2$ is used) and $U$ is connected (again, $n\ge2$) and orientable, so this determines a unique orientation of $U$ compatible with the orientation on $M\setminus\varphi(\mathrm{int} B^n)$. In the same manner, but reverse, the intersection $U\cap M^{\prime}\setminus\varphi^{\prime}(\mathrm{int} B^n)$ is connected and $M^{\prime}\setminus\varphi^{\prime}(\mathrm{int} B^n)$ is connected (using $n\ge2$ again) and orientable, so this determines a unique orientation of $M^{\prime}\setminus\varphi^{\prime}(\mathrm{int} B^n)$ compatible with the orientation on $U$. These compatible orientations on $M\setminus\varphi(\mathrm{int} B^n)$, $U$ and $M^{\prime}\setminus\varphi^{\prime}(\mathrm{int} B^n)$ glue together to a unique orientation of $M\#M^{\prime}$, which is compatible with the orientation on $M\setminus\varphi(\mathrm{int} B^n)$ coming from the orientation on $M$. This answers questions (1) and (2) affirmatively.
To answer questions (3) and (4), we conduct a more explicit analysis of the above. Note that the uniqueness claim of the previous paragraph means we only have to understand what orientation the unique orientation of $M\#M^{\prime}$ induced by the orientation on $M\setminus\varphi(\mathrm{int} B^n)$ induces on $M^{\prime}\setminus\varphi^{\prime}(\mathrm{int} B^n)$. Choose a collar $c\colon[0,1)\times S^{n-1}\rightarrow M\setminus\varphi(\mathrm{int} B^n)$, which coincides with the given embedding $\varphi\vert_{S^{n-1}}$ at the boundary. The fact that $\varphi\colon B^n\rightarrow M$ is orientation-preserving means that $\varphi\vert_{S^{n-1}}\colon S^{n-1}\rightarrow\partial(M\setminus\varphi(\mathrm{int} B^n))$ is orientation-reversing if the codomain is equipped with the boundary orientation (since $\varphi$ pushes forward inward-pointing tangent vectors along $S^{n-1}$ to outward-pointing tangent vectors along $\partial(M\setminus\varphi(\mathrm{int} B^n))$). This in turn implies that the collar $c$ is orientation-preserving (since the positively oriented tangent vector to $[0,1)$ at $0$ pushes forward to an inward-pointing tangent vector along $\partial(M\setminus\varphi(\mathrm{int} B^n))$).
Next, choose a collar $c^{\prime}\colon(-1,0]\times S^{n-1}\rightarrow M^{\prime}\setminus\varphi^{\prime}(\mathrm{int} B^n)$. The two collars glue together to the bicollar $C\colon(-1,1)\times S^{n-1}\stackrel{\sim}{\rightarrow}U\subseteq M\#M^{\prime}$. If we now follow the construction from earlier explicitly, the orientation on $M$ determines the orientation of $U$ obtained by pushing forward the standard orientation of $(-1,1)\times S^{n-1}$ along $C$, by the previous paragraph. This in turm means the compatible orientation on $M^{\prime}\setminus\varphi^{\prime}(\mathrm{int} B^n)$ is uniquely determined by making $c^{\prime}$ orientation-preserving. Reversing the arguments from the previous paragraph, this is equivalent to $\varphi^{\prime}$ being orientation-reversing (note crucially the difference between one collar neighborhood defined on $[0,1)\times S^{n-1}$ and the other on $(-1,0]\times S^{n-1}$). From this, we obtain that two orientations on $M$ and $M^{\prime}$ induce an orientation on $M\#M^{\prime}$, agreeing with the given ones on $M\setminus\varphi(\mathrm{int} B^n)$ and $M^{\prime}\setminus\varphi^{\prime}(\mathrm{int} B^n)$ respectively, if and only if the embedded discs are oppositely oriented.
I want to stress that this is the expected behavior. To illustrate this, let's think of $\mathbb{R}^n$ as the union of upper half-space $\mathbb{H}_+^n$ and lower half-space $\mathbb{H}_-^n$, which is topologically the local model for gluing manifolds along their boundaries. The intersection $\mathbb{H}_+^n\cap\mathbb{H}_-^n=\partial\mathbb{H}_+^n=\partial\mathbb{H}_-^n$ is a hyperplane and reflection along this hyperplane is a homeomorphism between $\mathbb{H}_+^n$ and $\mathbb{H}_-^n$, which is the identity on the boundary, but it reverses orientations! So if you want to glue upper and lower half-space along their boundary to get $\mathbb{R}^n$ in the oriented category, they need to be oppositely oriented. Equivalently, you glue them along an orientation-reversing diffeomorphism. This manifests in the above proof, which is basically noting that if you identify the boundaries in the given manner, what looks outward-pointing in one gluing component has to look inward-pointing in the other and vice versa.
Edit: Ok, so if you have embeddings $\varphi\colon B^n\rightarrow M$ and $\varphi^{\prime}\colon B^n\rightarrow M^{\prime}$, let $M\#_{(\varphi,\varphi^{\prime})}M^{\prime}$ denote the specific connected sum constructed from this data. Palais' theorem implies that if $\psi\colon B^n\rightarrow M$ and $\psi^{\prime}\colon B^n\rightarrow M^{\prime}$ are two more embeddings, such that $\varphi,\psi$ and $\varphi^{\prime},\psi^{\prime}$ are respectively equioriented, then $M\#_{(\varphi,\varphi^{\prime})}M^{\prime}\cong M\#_{(\psi,\psi^{\prime})}M^{\prime}$. Thus, we can write $M\#_{(\pm,\pm)}M^{\prime}$, where the signs denote if the respective embeddings preserves or reverses orientations, and obtain 4 well-defined diffeomorphism types. Now if $(\varphi,\psi)$ is a pair of embeddings, then $(\varphi r,\psi r)$, where $r\colon B^n\rightarrow B^n$ is a reflection (so orientation-reversing), is a pair of embeddings whose signs are opposite and you can see immediately that $M\#_{(\varphi,\psi)}M^{\prime}$ and $M\#_{(\varphi r,\psi r)}M^{\prime}$ are described by the same equivalence relation on the same space, hence equal. Thus, there really are only two diffeomorphism types worth considering, $M\#_+M^{\prime}$ and $M\#_-M^{\prime}$, where the sign now denotes whether these are realized by a pair of embeddings that have the same or the opposite effect on the respective orientation. Conventionally, one writes $M\#M^{\prime}$ to mean the diffeomorphism type $M\#_-M^{\prime}$, because this is the one that comes with a natural orientation compatible with both $M$ and $M^{\prime}$, as discussed previously.
These two diffeomorphism types can be genuinely different. The standard example is that $\mathbb{CP}^2\#\mathbb{CP}^2$ and $\mathbb{CP}^2\#-\mathbb{CP}^2$ have different signatures, so they aren't even homeomorphic. However, if $M$ has an orientation-reversing diffeomorphism $\alpha\colon M\rightarrow M$ and $(\varphi,\psi)$ realizes $M\#_-M^{\prime}$, then $(\alpha\varphi,\psi)$ realizes $M\#_+M^{\prime}$ and you can check that the diffeomorphism $M\setminus\varphi(\mathrm{int} B^n)\rightarrow M\setminus\alpha\varphi(\mathrm{int} B^n)$ induced by $\alpha$ glues together with the identity on $M^{\prime}\setminus\varphi^{\prime}(\mathrm{int} B^n)$ to yield a diffeomorphism $M\#_-M^{\prime}\cong M\#_+M^{\prime}$. The same argument works with the factors interchanged, of course. In total, we obtain that if one of $M$ or $M^{\prime}$ possesses an orientation-reversing diffeomorphism, there is only one diffeomorphism type of manifolds that can be obtained from connected sum constructions, but if neither of them has an orientation-reversing diffeomorphism, you can get two distinct diffeomorphism types (I do not know if they always are).
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God, it's late and I really shouldn't have asked this question. There's an easy counterexample: Take the Möbius strip / a rectangle whose vertical sides are identified and carry out two vertical cuts, so that it becomes a piecewise smooth manifold with two pieces, each of which is going to be orientable.