This is not a question about two diffeomorphisms composition.
Consider $\Phi : \Omega \to \widetilde\Omega$, where $\Phi$ is diffeomorphism, $\Omega$ and $ \widetilde\Omega$ are open subspaces of $\mathbb{R}^n$. I want to proof that $\forall a \in \Omega$, $\forall k \in \mathbb{N} < n$ there exist $U_a \subset \Omega$ and diffeomorphisms $\Phi_1: U_a \to \Phi_1(U_a), \Phi_2: \Phi_1(U_a) \to \widetilde\Omega$, such that $\Phi |_{U_a} = \Phi_2 \circ \Phi_1$ and $\Phi_1$ remains $k$ coordinates, while $\Phi_2$ remains $n - k$ other coordinates.
What I've done already: Let us consider $x \in \mathbb{R^k}, y \in \mathbb{R^{n - k}}$. Rewrite $\Phi(x, y) = (f(x, y), g(x, y))$ in the same manner. Now we want that $\Phi_1(x, y) = (x, \phi(x, y)), \Phi_2(x, y) = (\psi(x, y), y)$. Let's write a composition of them: $(f(x, y), g(x, y)) = \Phi(x, y) = \Phi_2 \circ \Phi_1(x, y) = \Phi_2(x, \phi(x, y)) = (\psi(x, \phi(x, y)), \phi(x, y))$. Now we can observe that $\phi = g$. What's left is to show that there exists $\psi$ such that $\psi(x, \phi(x, y)) = f(x, y)$. Note that it's eqivalent to $f = \psi \circ \Phi_1$. Now we want to apply the inverse function theorem: $\psi = f \circ \Phi_1^{-1}$. Can you help me to proceed?
Since everyone was busy stating "it's very much unclear", I decided to post the end of proof.
So, to apply inverse function theorem on $\Phi_1$, we need it to be continuously differentiable and it's Jacobian at point a non zero (invertible).
Differentiability (continuous) follows from differentiability of identity map $x \to x$ and f. Now let's write a Jacobian \begin{pmatrix} E & 0\\ f_x' & f_y' \end{pmatrix}. It is equal to $det(f_y')$. Note that it is a k-sized minor of $\Phi$. Since $\Phi$ is diffeomorphism, det$\Phi' \neq 0$ and there exist some k-sized non-zero minor. Now we can swap the coordinates as we want the right bottom minor to be non-zero.