I try to prove
For any matrixes $A_{ms},B_{sn}$ $$\operatorname{rank}{A}+\operatorname{rank}{B}-s\leq\operatorname{rank}{AB}$$
First, as for any $X$ that $BX=0$ also $ABX=0$, that $$\operatorname{dim}\operatorname{ker}B\leq\dim\ker{AB}$$
But second I have other idea. By def, kernel of $A$ is all $X$ from $\mathbb{R}^{s}$ that $AX=0$. Now, we can look at $ABX=0$ as $A(BX)=0$, so we can apply $A$ only on images of $B$. Thus, I conclude that any $X$ that is in kernel of $AB$ has to be in $\operatorname{ker}A$ and to be in $\operatorname{Im}B$: $\operatorname{ker}A\cap\operatorname{Im}B$. So
$$\operatorname{dim}\operatorname{ker}AB\leq\dim\ker{A}$$
But it isn't true, as we can input $A=E$ for example.
Please, help me to catch errors in my second statement and give a hint to prove statement in blockquote.
One problem with your argument is that an $X\in \operatorname{ker}(AB)$ is in particular an element of $\mathbb{R}^n$. Thus it makes no sense, to talk about whether it is in $\operatorname{ker}(A)\cap \operatorname{Im}(B)$. Since $B$ is not necessarily injective, it also does not correspond uniquely to an element of $\operatorname{Im}(B)$.
For a hint, how to prove the statement. Note that $\operatorname{rank}(AB)=\dim\operatorname{Im}(AB)$ and use the dimension formula on $AB$ as well as on $\tilde{A}\colon \operatorname{Im}(B)\to \mathbb{R}^m$. If you need a further hint, just comment.