Setup:
We note the following theorem: Let $U \subset \mathbb{R}^m$ and let $f \in C^{\infty}(U,\mathbb{R}^n)$. Now if $n \leq m$ and $Df|_x$ is surjective, there is a neighbourhood $V \subset \mathbb{R}^m$ of $x$ and a diffeomorphism $$\varphi:\mathbb{R}^m \supset W \to V$$ so that $$f \circ \varphi|_W = \operatorname{pr}:(x^1,\ldots,x^n,x^{n+1},\ldots,x^m) \mapsto (x^1,\ldots,x^n).$$
First a question related to this: Does $\mathbb{R}^{m} \supset W$ imply that we are taking $\mathbb{R}^m$ as domain of $\varphi$? I see no other choice since the restriction $\varphi|_W$ does not make sense otherwise.
Now we want to show that if for all $p \in N \subset M$ there exists a neighbourhood $V$ of $M$ (where $M$ is a smooth manifold) and a submersion $$f:V \to \mathbb{R}^{m-n}$$ such that $f^{-1}(0) = V \cap N$, then this implies that $N$ is a submanifold of codimension $m-n$.
Proof sketch:
Now I skip the details, but essentially we show that for a chart $(U,\varphi)$ of $M$ and $f$ as in the statement of the theorem, we have that $$f \circ \varphi^{-1}:\varphi(U) \subset \to \mathbb{R}^{m-n}$$ is a submersion. We can now use the theorem in the setup. Here is where I get confused; my teacher says that we can then find a neighbourhoods $\tilde{U} \subset U$ and $\tilde{W}$ so that $$\psi: \varphi(\tilde{U}) \to \tilde{W}$$ is a diffeomorphism and $$f \circ \varphi^{-1} \circ \psi^{-1} = \operatorname{pr}.$$
Now, first of, why not say ”we can find $\tilde{U} \subset \varphi(U) \subset \mathbb{R}^{m}$ s.t. etc” since I believe that is what the theorem says? But in one sense I believe this is equivalent since $\varphi$ is a diffeomorphism, for all open sets $A \subset \varphi(U)$ we must have that $\varphi^{-1}(A) = \tilde{U}$ is open in $U$. Second; according to the thereom, is it not that we should find a diffeomorphism $\psi$ so that $$f \circ \varphi^{-1} \circ \psi|_W = \operatorname{pr}?$$ I think my last question is not so important, maybe, since of course the inverse is also a diffeomorphism ($\langle \operatorname{Diff}(M),\circ \rangle$ forms a group).
Any clarification on how to think here would be helpful.