It's the statement of Implicit function theorem from Rudin's PMA. Why here consider mapping from $\mathbb{R}^{n+m}$ into $\mathbb{R}^{n}$? Why the dimension of domain is bigger than codomain? I guess that here $m\geqslant 1$.
I can't understand this moment
It is instructive to look at the statement of the theorem for $f$ a linear transformation $\mathbb{R}^{n+m} \to \mathbb{R}^n$ for different values of $m$. Note that the derivative of a linear map is itself, so the condition $A_x$ is invertible is the same as the upper left $n \times n$ minor of $f$ is invertible. In particular $f$ has rank $n$. By the rank-nullity theorem, we know that $\ker f$ is an $m$ dimensional linear subspace of $\mathbb{R}^{n+m}$.
Now switch back to more general $f$. The implicit function theorem tells us that if $f$ is $0$ at $(a,b)$, then the set $\{ f(x,y) = 0 : (x,y) \in \mathbb{R}^{n+m} \}$ can be parameterized locally (in some suitably small open neighborhood) with $m$ variables. That is, it is locally an $m$-dimensional submanifold of $\mathbb{R}^{n+m}$. That is, it can be written locally as the graph of a function of $m$ variables.
You can view the implicit function theorem as recovering the nice fact from linear algebra (rank-nullity) for more general functions, but only in a small neighborhood of $(a,b)$.
Now let $m = 0$. We can apply the implicit function theorem all the same. (You can see that $y = \mathbb{R}^0 = \{ 0 \}$ and $g : \{ 0 \} \to \mathbb{R}^n$ has $g(0) = (a,b)$.) The theorem is then stating that the $0$-set of $f$ is locally a point, or a $0$-dimensional manifold, etc.. This is also very important! The inverse function theorem is an even better result in this case: it tells us that $f$ is locally invertible.
Rank-nullity shows that $\ker f = \{ 0 \}$ when $f$ is a linear transformation. This is just the statement that the $0$-set of $f$ is a point, but it is a global result (from the linearity of $f$). Again we see the same correspondence: the implicit function theorem is a local version of this result.
Finally let $m < 0$. Here we cannot apply the implicit function theorem as stated. The more general constant rank theorem applies. The analog in this case is: if the derivative of $f$ is one-to-one at $(a,b)$, then $f$ is locally one-to-one. In particular the zero-set of $f$ is again locally just the point $(a,b)$.
We see that the number $m$ is the dimension of the zero-set of $f$ (locally). This is why the theorem is stated for $m \geq 0$.
I apologize if this answer is a little scattered/unfocused.
