Oriented Hypersurface admits unique unit normal vector field This question is the converse of this question and is taken from Lee's Smooth Manifolds, problem 15.7. Namely:

Suppose $M$ is an oriented Riemannian manifold and $S\subset M$ is an oriented smooth hypersurface. Show that there is a unique smooth unit normal vector field along $S$ that determines the given orientation of $S$.

So we can choose an orientation on $S$ via orthonormal coordinate charts and complete this to an orientation on $M$. However, this is in no way uniform, so we can't make it a vector field. Not sure where to go from here.
 A: I know this is an old post, but here is a detailed solution, which I wrote partly as an exercise for myself to go through the details.
First we define a rough unit vector field $N$ that determines the orientation on $S$. Let $p\in S$, and note that we can write $T_p M=T_pS\oplus (T_pS)^\perp$. Since $S$ is a hypersurface, we know that $(T_pS)^\perp$ is $1$-dimensional, and since $\Vert N_p\Vert=1$, there are two candidates for $N_p$. Consider an orientation form $\omega$ of $M$. For $N$ to determine the orientation, we need given an oriented basis $(E_1\vert_p,\dots,E_{n-1}\vert_p)$ of $T_pS$, that $\omega(N_p,E_1\vert_p,\dots,E_{n-1}\vert_p)>0$. Clearly there is only one choice for $N_p$. This choice is well-defined, for given any other oriented basis $(\tilde E_i\vert_p)$, we know the transition matrix between $(E_i\vert_p)$ and $(\tilde E_i\vert_p)$ has positive determinant, and therefore the choice of $N_p$ would be the same for $(\tilde E_i\vert_p)$.
We now show smoothness of $N$. It suffices to show this locally. Consider an oriented orthonormal frame $(E_1,\dots,E_{n-1})$ on a neighbourhood $U\ni p$ of $S$. We can think of these as maps
$$
U\to TS\vert_U\to TM\vert_U.
$$
Consider $N_p\in(T_pS)^\perp$. Note that
$$
N_p=N_p-\sum_{i=1}^{n-1}\langle N_p,E_i\vert_p\rangle.
$$
We can extend $N_p$ locally to a vector field $X$ and define $\tilde N$ by
$$
\tilde N=X-\sum_i^{n-1}\langle X,E_i\rangle.
$$
By the observation above $N_p=\tilde N_p$, so $\omega(\tilde N_p,E_1\vert_p,\dots,E_{n-1}\vert_p)>0$, and therefore by continuity of $\omega$ we also have $\omega(\tilde N,E_1,\dots,E_{n-1})>0$ locally. It follows that $N=\tilde N$ locally, and therefore $N$ is smooth.
