Is the statement true ? Let's suppose $M,N$ be smooth-manifolds without boundary for the moment, I think I got the proof if I'm able to find $y \in \text{RegVal}(f) \cap \text{Im}(f)$ arguing as follows : by the sake of contradiction let's $f$ be injective, then $f^{-1}(y) = \left\lbrace x\right\rbrace$, but for a known theorem $f^{-1} (y)$ is an $(m-n)$ manifold.
Since $m > n$, we have that this can't happens since $\left\lbrace x\right\rbrace$ is a $0-$manifold.
This seems correct, but I can always find some $y \in \text{RegVal}(f) \cap \text{Im}(f) ?$ And what if boundaries are allowed ?
Any help would be appreciated
No, you most certainly cannot always find a regular value in the image of $f$. Consider the map $\Bbb R^m\to\Bbb R^n$ (with $m>n$) given by $$f(x_1,\dots,x_n,x_{n+1},\dots,x_m) = (x_1,\dots,x_{n-1},0).$$ That said, this map has constant rank and certainly will not be injective. In fact, even though $f$ may not be a submersion, in a neighborhood of a point of maximum rank the rank of $f$ will be constant and the map will, in appropriate coordinates, look like such a mapping. Thus, it fails to be injective in a neighborhood of any point of maximum rank. Where the rank drops, things get much more difficult. But this is still enough to answer your question.