Meant by a unit outer normal along an oriented hypersurface in an oriented Riemannian manifold

Question

I know that given a manifold $M$ with boundary ($\partial M\neq\emptyset$) and a point $p\in\partial M$, the vectors in $T_p M$ can be classified into three types: those which are also in $T_p(\partial M)$, those which are not in $T_p(\partial M)$ and point inward, and those which are not in $T_p(\partial M)$ and point outward. For the definition of the last two types, $v\in T_p M\setminus T_p(\partial M)$ is said to be inward-pointing if for some $\epsilon>0$, there exists a smooth curve $\gamma:[0,\epsilon)\to M$ such that $\gamma(0)=p$ and $\gamma'(0)=v$; it is said to be outward-pointing if there exists such a curve with its domain replaced by $(-\epsilon,0]$.

Now I'm wondering if we can proceed the above discussion by replacing $\partial M$ with any other hypersurface $S$ in $M$. To be precise, let us endow the manifold $M$ with a Riemannian metric $g$ and suppose $M$ is oriented. We also assume the hypersurface $S$ is oriented. Then, my question is, what could be a sensible interpretation if we came across the statement that $N$ is a unit outer normal along $S$?

Except for the outer part, I do know a little bit about the statement that $N$ is a unit normal along $S$, which means $N$ is a unit normal vector field along $S$, that is, a map $N$ assigns to each $p\in S$ a unit vector $N_p\in T_p M$ that is orthogonal to all vectors in $T_p S$, where $T_p S$ is identified with a subspace of $T_p M$ according to an usual convention. And since $T_p M=T_p S\oplus N_p S$ with $N_p S:=(T_p S)^\perp$, the nonzero vector $N_p$ should fall into $T_p M\setminus T_p S$. Now it seems that we can mimic the discussion in the first paragraph to define a unit outer normal, but I'm not sure. For what is worth, I know a theorem guarantees that there exists a unique unit normal vector field along $S$ that determines the given orientation of $S$.

Does anyone know the specific definition of a unit outer normal? I'd really appreciate it if you could kindly recommend me a textbook that clearly defines the term. Thank you.

Edit. I'm going to include the passage where I found the statement in question. The passage came from Dan Lee's AMS textbook Geometric Relativity:

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Answer 1

• For a general hypersurface $S$, no. The adjectives ‘outer/inner’ only make sense for boundaries of manifolds. These adjectives make sense here (just from a linguistic perspective) because $n$-dimesnional manifolds-with-boundaries are modelled on the closed half-spaces $\Bbb{H}^n_+$, where along the boundary we have an obvious notion of which vectors point ‘inward’ vs ‘outward’.
• If $S$ is a general hypersurface, and $M,S$ have nice topological properties, then there exist exactly two unit normal vector fields defined on $S$. If you call one of them $N$, the other is $-N$; but there is no reason to call one ‘outer’ and the other ‘inner’. We simply say “orient the hypersurface by the unit normal $N$” (or $-N$ if that’ what you want).

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The author is not ascribing the phrase ‘outer’ to an arbitrary $\Sigma$. In the penultimate sentence he says “… and $\Sigma$ has an inside and outside…”. The only way I make sense of this is to suppose you have a hypersurface $\Sigma$ such that your manifold $M$ is a disjoint union of three sets $\Omega_0,\Sigma,\Omega_1$ such that $\Omega_0,\Omega_1$ are open sets, and $\widetilde{\Omega}_0:=\Omega_0\cup\Sigma$ and $\widetilde{\Omega}_1:=\Omega_1\cup\Sigma$ are both smooth manifolds-with-boundaries having $\Sigma$ as their boundary and that one of them, say $\widetilde{\Omega}_0$ is compact. Then, the normal you should choose on the hypersurface $\Sigma$ is the outward normal of the compact manifold-with-boundary $\widetilde{\Omega}_0$.

If you think about my last comment I made, this means spheres should be oriented by outward normal of the closed ball. Why? Probably because in the long run, it will save us several minus signs with the divergence theorem.