Recover the differential form by its pullback I have a smooth manifold $M$ in $\mathbb{R}^n$ given by $M = \{ x \in \mathbb{R}^n \mid g(x) = 0 \}$. Its atlas consists of a single chart $(M,\varphi)$, where
$$
\begin{array}{rcl}
\varphi(x_1,...,x_n) & = & (x_2,...,x_n), \\
\varphi^{-1}(y_1,...,y_{n-1}) & = & (x_{1}(y),y_1,...,y_{n-1})
\end{array}
$$
where $g(x_{1}(y),y_1,...,y_{n-1}) = 0$ (such $x_{1}(y)$ is unique) and $\varphi(M) = \mathbb{R}^{n-1}$.
A differential form $\omega$ on $\mathbb{R}^{n-1}$ is given by
$$
  \omega = \frac{a(x_{1}(y),y_1,...,y_{n-1})}{\partial_{1} g(x_{1}(y),y_1,...,y_{n-1})}  dy_{1} \wedge ... \wedge dy_{n-1}.
$$
Is it possible to find a form $\alpha$ on $M$ such that $\omega = (\varphi^{-1})^{*} \alpha$?
 A: Assuming the map $\varphi^{-1}:\mathbb{R}^{n-1}\to\mathbb{R}^n$ is smooth (which, if you are defining $\varphi$ the way I think you are, should be automatic). 
Then by definition $\mathrm{d}\varphi^{-1}: T_x\mathbb{R}^{n-1} \to T_pM$ where $\varphi(p) = x$. To find an $\alpha$ defined along $M$ such that it pullsback via $\varphi^{-1}$ to $\omega$, it suffices to specify, since $\alpha(p) \in \wedge^{n-1} T_p^*\mathbb{R}^n$, how it acts on a basis of $\wedge^{n-1} T_p\mathbb{R}^n$. 
For the following we´ll also need to assume that $\iota := \mathrm{d}\varphi^{-1}$ is injective. Let $e_i$ be the standard basis vector fields on $\mathbb{R}^{n-1}$. Then by construction $f_i = \iota\circ e_i$ are linearly independent vector fields along $M$ that span $TM$. Smoothly complete this with a smooth vector field $f_0$ along $M$ that is transversal to $TM$, then $\{f_0, f_1,\ldots, f_{n-1}$ define a basis for $T\mathbb{R}^{n}$, which induces a basis for $\wedge^{n-1}T\mathbb{R}^n$. 
Now define $\alpha$ such that $\alpha(f_i\wedge f_j \wedge \cdots \wedge f_k) = \omega(e_i\wedge e_j\cdots\wedge e_k)$ if none of the indices are 0. And define $\alpha(f_0\wedge\cdots) = 0$. It is easy to check that by definition $\omega = (\varphi^{-1})^* \alpha$. 
Notice that the $\alpha$ depends on how you ¨complete the basis¨, that is, which vector field you choose as $f_0$. But the restriction of all the different $\alpha$s to $TM$ are the same, since $f_1, \ldots, f_{n-1}$ span $TM$. Notice that if $\alpha$ and $\alpha´$ are two such forms, we have that $(\alpha - \alpha´)(X_i\wedge\cdots\wedge X_j) = 0$ whenever all $X_i\in TM$. By dimension counting and a bit of linear algebra, you´d see that $\wedge^{n-1}T\mathbb{R}^n$ is $n$ dimensional, and so the space of $\beta$s with the property that it vanish on $(X_i \wedge \cdots \wedge X_j)$ for $X_i\in TM$ is $n-1$ dimensional. Since the image of $dg \wedge : \wedge^{n-2}T\mathbb{R}^n \to \wedge^{n-1}T\mathbb{R}^n$ is also $n-1$ dimensional, you have that necessarily $\alpha - \alpha´ = dg \wedge \gamma$ for some $n-2$ form $\gamma$. 
