The situation: Let $Σ^{n-1} ⊂ ℝ^n$, $n ≥ 2$, be an oriented, compact, embedded smooth hypersurface with boundary. For simplicity assume $∂Σ = \{ x ∈ ℝ^n \,|\, (x¹)² + … + (x^{n-1})² = R², x^n = 0 \}$, i.e. $∂Σ$ is the $(n-2)$-sphere of radius $R$ in the $x^n = 0$ plane. Consider the restriction of the (spherical) radius function on $ℝ^n$ to $Σ$ and let $r < R$ be a regular value with non-empty preimage. Put differently, $Σ$ and the $(n-1)$-sphere $∂B_r(0)$ intersect transversally, i.e. their intersection $I$ is a smooth compact $(n-2)$ submanifold which is canonically oriented as a level set of a function on the oriented $Σ$ (consider the gradient).
Clearly, in $n = 2$, the intersection $I$ will just be a finite union of points and in $n = 3$ dimensions, it will be a finite union of smooth Jordan curves. More generally, for $n \geq 3$, it will be the finite union of closed connected and oriented $(n-2)$-surfaces, each of which separates $∂B_r(0)$ into two disjoint open sets by the Jordan-Brouwer separation theorem. (See this question for a version that applies to $S^{n-1}$, where $n \geq 3$.)
The goal: I'm interested in cutting out the open ball $B_r(0)$ from $Σ$ and gluing in, in its stead, a piece $Ω$ of $∂B_r(0) ≌ S^{n-1} $ in such a way that the resulting manifold $Σ' ≔ (Σ ∖ B_r(0)) ∪ Ω$ is a piecewise smooth hypersurface which has the same boundary as $Σ$ and which is orientable(★) and, in fact, canonically oriented in such a way that the orientations of $Σ$ and $Σ'$ agree on their intersection.
Side note (★): By an orientation of a piecewise smooth hypersurface $Σ'$ I mean an orientation of the individual smooth pieces in such a way that any two adjacent pieces $P_1, P_2$ induce on their common edge $E$ opposite orientations. (So that if I take an exact $(n-1)$-form $dω$ on $ℝ^n$ and integrate it over $Σ'$, Stokes' theorem will not give a contribution for the edges and only the value of $ω$ on $∂Σ'$ will be relevant.)
Attempt at a solution: The detailed proof of finding a candidate for $Ω$ is a bit technical but, essentially, it is just an iteration over the connected components $Z$ of $∂B_r(0) ∖ I$, where one "colors" adjacent components red and blue in an alternating fashion and, later, chooses $Ω$ to be the union of either all red components or all blue components. The resulting $Σ'$ will then have the same boundary as $Σ$ as all the boundaries of $Σ ∖ B_r(0)$ on $∂B_r(0)$ have been "closed".
My only problem is the orientability of $Σ'$. From various drawings in 2 and 3 dimensions I've managed to convince myself that $\Omega$ can always be oriented in the right way, i.e. so that $\Sigma'$ is oriented (see an example for $n = 2$ below), but I can't come up with a good (general) reason as to why that is.
$n=2$ dimensions">Heuristically speaking, the induced orientations on "nearby" connected components of the intersection $I$ are opposite because when $Σ$ enters the ball, it also has to leave it at some point. But obviously this is not a proof.
Is this in some way related to intersection numbers? My differential topology is a bit rusty, so I'd appreciate answers at a beginner's level (if there are any). Also, I would strongly prefer answers that don't use the topology of the inside of the ball $B_r(0)$ as in the actual case I'm interested in, I'm dealing with an ambient manifold $M$ which is only asymptotically flat, i.e. there is a compact set $K ⊂ M$, such that $M ∖ K$ is diffeomorphic to $ℝ^n ∖ B_\rho(0)$, where $\rho < r$. Put differently, the ambient manifold might have non-trivial topology somewhere inside the ball $B_r(0)$ from above.