In my quest to find an answer to this question of mine, I have reduced the setup to the following conjecture:
Let $M^n$ ($n ≥ 2$) be an oriented smooth manifold, $Σ^{n-1} ⊂ M$ be a connected, oriented, embedded smooth hypersurface (possibly with boundary) and $S ⊂ M$ be a smooth embedded hypersurface which is homeomorphic to the sphere $S^{n-1}$. Suppose $S$ separates $M$ into two disjoint open sets (i.e. $M ∖ S$ has two connected components) and call one of them $B$. Suppose furthermore that $Σ$ intersects $S$ transversally and that $Σ ∩ S$ has precisely two connected components $γ_1$ and $γ_2$ (which are automatically compact $(n-2)$-dimensional submanifolds). Then, by the Jordan-Brouwer separation theorem for the sphere, each $γ_i$ separates $S$ into two disjoint open sets, say $V_i$ and $W_i$. W.l.o.g. assume that $W_1$ contains $γ_2$ and $W_2$ contains $γ_1$. Let $U ≔ W_1 ∩ W_2$ and $V ≔ V_1 ∪ V_2$. Clearly, the topological boundary of both $U$ and $V$ is $∂U = γ_1 ⊍ γ_2 = ∂V$.
Consider $Σ ∖ B$ which is a smooth hypersurface with boundary $γ_1 ⊍ γ_2$. It induces on each $γ_i$ an orientation. Call this orientation $$.
Claim: The orientations on the manifolds with boundary $\bar{U} = U ∪ ∂U$ and $\bar{V} = V ∪ ∂V$ can be chosen in such a way that $\bar{U}$ induces on both $γ_i$ the opposite orientation $-$ and similarly for $\bar{V}$. Put differently, with this choice of orientation on $U$ and $V$, both $(Σ ∖ B) ∪ \bar{U}$ and $(Σ ∖ B) ∪ \bar{V}$ will be oriented piecewise smooth submanifolds. (Here, an oriented piecewise smooth manifold is one where every piece is an oriented smooth manifold-with-boundary and neighboring pieces induce opposite orientations on their common edges / boundary components, so that Stokes' theorem will not give a contribution along these edges.)
Similarly, flipping the orientations on $U$ and $V$ will make $(Σ ∩ B) ∪ \bar{U}$ and $(Σ ∩ B) ∪ \bar{V}$ oriented piecewise smooth submanifolds.
I'm mostly interested in proving the case where $M$ is asymptotically flat (i.e. $M ∖ K ≌ ℝ^n ∖ \overline{B_½(0)}$ outside a compactum $K \subset M$) and $S ⊂ ℝ^n ∖ \overline{B_½(0)}$ is the usual sphere and $Σ$ is a compact hypersurface with boundary $∂Σ ⊂ M ∖ (B ∪ S)$. In this situation, I plan on gluing $Σ ∖ B$ and $Ω ≔ \bar{U}$ (or, alternatively, $Ω ≔ \bar{V}$) together to an oriented piecewise smooth hypersurface. (See my original question.)
Some thoughts:
Once $Σ ∖ B$ has induced an orientation $$ on its boundary components $γ_1$ and $γ_2$, the orientation on $\bar{U}$ (or $\bar{V}$) can be chosen in such a way that its boundary piece $γ_1$ has orientation $-$. Then the only thing that remains to show is that $\bar{U}$ also induces the orientation $-$ on $γ_2$. I first thought one might be able to use Stokes to deduce a contradiction otherwise but so far I don't see how.
Connectedness of $Σ$ seems crucial. Without it, counterexamples can be found easily.
Does the proof go through even when $S$ is not a sphere but any orientable, (connected), closed, embedded hypersurface which separates $M$?
I would be grateful for any ideas!