Assuming that my [edits to your question](https://mathoverflow.net/suggested-edits/27393) indeed reflect your intent correctly, the answer seems simple (and, indeed, was basically already sketched by Ville Salo in [a comment](https://mathoverflow.net/posts/comments/1102913) above).

First, let's assume that the local rules $f$ and $g$ for the original cellular automata have the same neighborhood size, i.e. that $m = n$.  (If that's not the case, you can always increase the size of the smaller neighborhood to make them equal.)

Now, we make use of the fact that the sets $A^{\mathbb Z} \times B^{\mathbb Z}$ and $(A \times B)^{\mathbb Z}$ are naturally isomorphic, with the isomorphism mapping any pair of configurations $(x_i)_{i \in \mathbb Z} \in A^{\mathbb Z}$ and $(y_i)_{i \in \mathbb Z} \in B^{\mathbb Z}$ to the configuration $(x_i, y_i)_{i \in \mathbb Z} \in (A \times B)^{\mathbb Z}$. (Of course, this works just as well with any index set, not just $\mathbb Z$.)

Thus, we can simply represent the product automaton as $(C^{\mathbb Z}, H)$, where the alphabet $C = A \times B$ of the product automaton consist of pairs of symbols $(x, y)$, with $x \in A$ and $y \in B$, and where the global rule $H$ is induced by the local rule $h: C^n \to C$ defined as: $$h((x_1, y_1), …, (x_n, y_n)) = (f(x_1, …, x_n), g(y_1, …, y_n)).$$

That is, to evaluate the local rule $h$, we take the $n$-letter word $((x_1, y_1), …, (x_n, y_n)) \in C^n$, unpack it into the words $(x_1, …, x_n) \in A^n$ and $(y_1, …, y_n) \in B^n$, pass those into $f$ and $g$ respectively and return the pair of resulting symbols.