How do we need to apply the inverse function theorem here? (diffeomorphically mapping an open subset of a submanifold)

I guess the desired claim follows from the following result:

Let $d_i\in\mathbb N$, $k_i\in\{1,\ldots,d_i\}$, $M_i$ be a $k_i$-dimensional embedded $C^1$-submanifold of $\mathbb R^{d_i}$ and $f:M_1\to M_2$ be $C^1$-differentiable at $x_1\in M_1$.

Assuming that $T_{x_1}(f):T_{x_1}\:M_1\to T_{f(x_1)}\:M_2$ is injective and $k:=k_1=k_2$ (if I'm not missing something, both assumptions together are equivalent to require that $T_{x_1}(f)$ is an isomorphism), we are able to apply the inverse function theorem and show that $\left.f\right|_{\tilde\Omega_1}$ is a $C^1$-diffeomorphism from $\tilde\Omega_1$ to $f(\tilde\Omega_1)$ for some $M_1$-open neighborhood $\tilde\Omega_1$ of $x_1$.

In fact, let $x_2:=f(x_1)$ and $\Omega_2$ be an $M_2$-open neighborhood of $x_2$. Since $f$ is continuous at $x_1$, $$f(\Omega_1)\subseteq\Omega_2\tag1$$ for some $M_1$-open neighborhood $\Omega_1$ of $x_1$. Since $M_i$ is a submanifold, there is a $C^1$-diffeomorphism $\phi_i$ from $\Omega_i$ to $U_i:=\phi(\Omega_i)$. Let $$\psi:=\phi_2\circ\phi_1^{-1}$$ and $u_i:=\phi_i(x_i)$. By assumption, ${\rm D}\psi(u_1)$ is injective (if I'm right with my remark above, it is even bijective) and hence, by the inverse function theorem, $\left.\psi\right|_{V_1}$ is a $C^1$-diffeomorphism from $V_1$ to $V_2$ for some $\mathbb R^k$-open neighborhood $V_1\subseteq U_1$ of $u_1$ and a $\mathbb R^k$-open neighborhood $V_2\subseteq U_2$ of $\psi(u_1)=\phi_2(f(x_1))$. Thus, $$\left.f\right|_{\tilde\Omega_1}=\phi_2^{-}\circ\left.\psi\right|_{V_1}\circ\left.\phi_1\right|_{\tilde\Omega_1}\tag2$$ is a $C^1$-diffeomorphism from $\tilde\Omega_1:=\phi_1^{-1}(V_1)$ onto $$\tilde\Omega_2:=\phi_2^{-1}(\underbrace{\psi(V_1)}_{=\:V_2})=f(\tilde\Omega_1)\tag3.$$ Since $\phi_i$ is continuous and $V_i$ is $U_i$-open, $\tilde\Omega_i$ is $\Omega_i$-open and hence a $M_i$-open neighborhood of $x_i$.