How to prove and interpret $\operatorname{rank}(AB) \leq \operatorname{min}(\operatorname{rank}(A), \operatorname{rank}(B))$?

Yes. If you think of A and B as linear maps, then the domain of A is certainly at least as big as the image of B. Thus when we apply A to either of these things, we should get "more stuff" in the former case, as the former is bigger than the latter.

Once you have proved $\operatorname{rank}(AB) \le \operatorname{rank}(A)$, you can obtain the other inequality by using transposition and the fact that it doesn't change the rank (see e.g. this question).

Specifically, letting $C=A^T$ and $D=B^T$, we have that $\operatorname{rank}(DC) \le \operatorname{rank}(D) \implies \operatorname{rank}(C^TD^T)\le \operatorname{rank} (D^T)$, which is $\operatorname{rank}(AB) \le \operatorname{rank}(B)$.

Prove first that if $f:X\to Y$ and $g:Y\to Z$ are functions between finite sets, then $|g(f(X))| \leq \min \{ |f(X)|, |g(Y)| \}.$

Then use the same idea.