How does a transition function behave in a FSM when an input is not recognized (i.e. the input is not contained in the alphabet)?
In my copy of the book, third edition, this is Theorem 1.25. From its proof, I quote:

The theorem remains true if they have different alphabets, $\Sigma_1$
and $\Sigma_2$. We would then modify the proof to let $\Sigma = \Sigma_1\cup \Sigma_2$.

What is suggested here is that in advance we assume the two alphabets to be equal, i.e., before we start the product construction. So we have to transform the original automaton $M_1$ with alphabet $\Sigma_1$ into an equivalent automaton $M_1'$ with alphabet $\Sigma_1\cup \Sigma_2$ accepting the same language. As you observe automata here are deterministic, i.e., have a transition function $\delta:Q\times \Sigma\to Q$.
The easiest way for the construction of $M_1'$ is extending the transition function for letters outside $\Sigma_1$. Any such letter should lead to not accepted computations. This can be done by adding a new garbage (or failure state) where all letters from $\Sigma_2\setminus \Sigma_1$ are mapped to.