How to show that $\dim\ker(AB) \le \dim \ker A + \dim \ker B $?

This is a proof in general where $A:V\to W$ and $B:U\to V$ are linear maps. Here $U$, $V$, and $W$ are arbitrary vector spaces over a base field $F$, and they do not necessarily have finite dimensions. That is, $$\dim \ker (AB) \leq \dim \ker A+\dim\ker B$$ is true whether or not the relevant dimensions are finite cardinals.

Note that $x\in \ker(AB)$ iff $Bx\in \ker A$, which is the same as saying $$x\in B^{-1}(\ker A\cap \operatorname{im}B).$$ Recall from the isomorphism theorems that $\operatorname{im} B\cong U/\ker B$ so there exists an isomorphism $$\varphi: U\overset{\cong}{\longrightarrow} \ker B\oplus \operatorname{im}B.$$ In other words, $$\varphi\big(B^{-1}(\ker A\cap \operatorname{im}B)\big)=\ker B\oplus (\ker A\cap \operatorname{im}B).$$ Consequently, \begin{align}\dim\ker(AB)&=\dim\big(\ker B\oplus (\ker A\cap \operatorname{im}B)\big)\\&=\dim\ker B+\dim(\ker A\cap \operatorname{im}B).\end{align} Since $\ker A\cap \operatorname{im}B\subseteq \ker A$, we obtain the desired inequality.