This question goes on where this question ended. Given is an non-empty $(n-1)$-dimensional ($n\ge2$) differentiable submanifold $X\subset\mathbb{R}^n$ such that there exists an open $U\subset\mathbb{R}^n$ with $X\subset U$ and a differentiable function $f:U\to\mathbb{R}$ such that $X=f^{-1}(0)$ and $Df(p)\neq0$ for all $p\in X$. Then $X$ is (as shown in the other question) the topological boundary in $U$ of $A:=\{p\in U\,|\,f(p)<0\}$.
I have to show that $X$ is an orientable manifold. That is, there is an differentiable atlas $\mathcal{A}$ such that $\det(D(h_i\circ h_j^{-1})(h_j(p))>0$ for all $p\in U_i\cap U_j$, for all charts $(U_i,h_,V_i),(U_j,h_j,V_j)\in\mathcal{A}$. Alternatively, there exists a $(n-1)$-form on $X$ which vanishes nowhere.
Now intuitively I feel like this is clear. Let $V\subset X$ be a connected component; I think that it is the case that $X$ is actually the smooth boundary of $A$, since if it wasn't, then $V$ (or some other connected component) would have some sort of kink somewhere, which would contradict it being a submanifold; as $A$ is open in $\mathbb{R}^n$ (since $U$ is open and $A$ is open in $U$), $A$ is orientable (restricting the standard orientation on $\mathbb{R}^n$ to $A$), and an orientation on $A$ induces an orientation on $\partial A=X$. Can this be turned into a formal argument?
Alternatively, how would one approach a proof using the two definitions of orientation given above? As the gradient $Df(p)$ is non-zero everywhere on $X$, and $X$ is the level set of $f$ at $C=0$, the gradient would be an ideal candidate to provide the outward pointing normal vector needed.
Any help is much appreciated. This question has kept me busy for some time now.
The following $n-1$ form $\omega$ on $X$ is nowhere zero and thus orients $X$:
For $x\in X$ and $v_1,v_2,\cdots,v_{n-1}\in T_x(X)=(\operatorname {grad}f(x) )^\perp\subset T_x(\mathbb R^n)= \mathbb R^n$ define $$\omega_x(v_1,v_2,\cdots,v_{n-1})=\det (v_1,\cdots,v_{n-1},\operatorname {grad}f(x) ).$$