Directly from Zorich, Mathematical analysis I, sec. 8.6, p. 505:
The question is very simple: the function $g:=f\circ \varphi^{-1}$ is a function of variables $(u_1,...,u_k)$ defined in an appropriate neighborhood of $\mathbb{R}^k \ni u_0=\varphi(x_0)$ while here it is used as if it were a function of the variables $ (y_1, ..., y_k) $. I can't understand how $ (u_1, ... , u_k)$ and $ (y_1, ..., y_k) $ can be interchangeable.
PS: As already done in other posts, I don't post screenshots of the whole proof of the theorem for copyright reasons, but I kindly ask anyone who has the book, if he can give me a hand in understanding.
This is indeed confusing but... it does make sense!!
$\bullet\enspace$ Because the functions we are dealing with have domain in the "standard" finite dimension space where there is a "canonical" coordinate, e.g. $$ g: \left\lbrace \begin{aligned} \tilde{O}(u_0)\subseteq \mathbb{R}^m\quad & \longrightarrow \quad \mathbb{R}^n\\ (u^1, \cdots, u^m) & \longmapsto \big(g^1(u^1, \cdots, u^m), \cdots , g^k(u^1, \cdots, u^m), g^{k+1}(u^1, \cdots, u^k),\cdots , g^{n}(u^1, \cdots, u^k) \big) \end{aligned} \right.$$ the $u^i$ are dummy variables.
(In fact you are correct, the first $k$ components of $g$ are just the "identity": $\forall\ j\in \mathbb{N},\ 1 \leq j \leq k,\ g^j(u^1, \cdots, u^m)= u^j$ so that $g$ really is a function of $(u^1, \cdots, u^k)$.)
So it does make sense to define a priori on the domain $\mathcal{D}:= \big(\tilde{O}(u_0)\cap \mathbb{R}^k)\times \mathbb{R}^{n-k}\enspace \subseteq \mathbb{R}^n$ $$ \psi: \left\lbrace \begin{aligned} \mathcal{D} \quad & \longrightarrow \quad \mathbb{R}^n\\ (u^1, \cdots, u^k, y^{k+1}, \cdots, y^n) & \longmapsto \big( u^1, \cdots , u^k, y^{k+1}- g^{k+1}(u^1, \cdots, u^k),\cdots , y^n- g^{n}(u^1, \cdots, u^k) \big) \end{aligned} \right.$$ The notation $\tilde{O}(u_0)\cap \mathbb{R}^k$ means the first $k$ components of the points in $\tilde{O}(u_0)$, i.e. its image by the projection on these components. It does make sense if one renames $y^i$ the dummy variables $u^i$ for $1\leq i \leq k$. The important thing is that there are the right number of variables.
$\bullet\enspace$ At this point of the proof, their is no constraint for the choice of a neighborhood $O(y_0)\subseteq \mathbb{R}^n$ of $y_0:= g(u_0)=\big(u_0^1, \cdots , u_0^k, g^{k+1}(u_0^1, \cdots, u_0^k),\cdots , g^{n}(u_0^1, \cdots, u_0^k) \big)$ and one checks that the previous choice $\mathcal{D}$ does contain $y_0$.