If $L$ is context-free then there is a PDA $\mathscr{P}$ that accepts it. If $M$ is regular then there is a DFA $\mathscr{F}$ that accepts it. The intersection language consists of the words that are recognized by $\mathscr{P}$ and $\mathscr{F}$.
Any word that is in the intersection is accepted by $\mathscr{F}$, but not all words that are accepted by $\mathscr{F}$ are in the intersection: only those that are also accepted by $\mathscr{P}$.
The cross product proof consists of constructing an automaton $\mathscr{P} \otimes \mathscr{F}$ which contains the mechanics of both $\mathscr{P}$ and $\mathscr{F}$, and which accepts only words for which both sides accept. The cross-product automaton is a PDA (and therefore the recognized language is context-free) — intuitively, because the cross product with an $n$-state DFA consists of taking $n$ copies of $\mathscr{P}$ and adding $(q,a,[q])$ arrows between matching states in $\mathscr{P}$ where the DFA has $a$ arrows. The result is not a finite automaton in general (not even a non-deterministic one) because the $\mathscr{P}$ part relies on the stack and this reliance does not go away in $\mathscr{P} \otimes \mathscr{F}$ in general.
A trivial example is that $\mathscr{A}^*$ is regular, and if $L$ is context-free but not regular then $L \cap \mathscr{A}^* = L$ is context-free but not regular.
