Apologies for the vague title, but I can't describe the following question well.
The question is (as usual, from my book)
Let f:X \to X' and g:Y \to Y' be continuous. If \mbox{cls} \varphi \in H^p(X';R) and \mbox{cls} \theta \in H^q(Y';R), then $(f \times g)^*(\mbox{cls} \varphi \times \mbox{cls} \theta) = f^* \mbox{cls} \varphi \times g^* \mbox{cls} \theta.$
(here $R$ is a commutative ring). So the way I read the question, the multiplication is the cup product. I will first introduce the notation used in the book - I am not sure if it is standard.
If $\varphi \in S^n(X,G)$ and $c \in S_n(X)$, we write $(c,\varphi)=\varphi(c) \in G.$
Let $X$ be space, and let $R$ be a commutative ring. If $\varphi \in S^n(X;R)$ and $\theta \in S^m(X;R)$, define the cup product $\varphi \cup \theta \in S^{n+m}(X,R)$ by $(\sigma,\varphi \cup \theta) = (\sigma \lambda_n, \varphi)(\sigma \mu_m,\theta)$ for every $(n+m)$-simplex $\sigma$ in $X$, where the right side is the product of two elements in the ring $R$. ($\mu$ and $\lambda$ are the back face and front face functions respectively)
So I started looking at $f^* \mbox{cls} \varphi \times g^* \mbox{cls} \theta$. We have that $ \begin{align} f^* \mbox{cls} \varphi \times g^* \mbox{cls} \theta &=\mbox{cls}(f^\# \varphi)\mbox{cls}(g^\# \theta) \\ &=\mbox{cls}([f^\# \varphi] \smile [g^\# \theta]) \\ &=\mbox{cls}\left((\sigma \lambda_p, f^\# \varphi)(\sigma \mu_q,g^\#\theta)\right) \\ &=\mbox{cls}\left((f \sigma \lambda_p,\varphi)(g \sigma \mu_q,\theta)\right) \end{align} $ The other side is just $ \begin{align} (f \times g)^*(\mbox{cls} \varphi \times \mbox{cls}\theta) &=(f \times g)^*\mbox{cls}(\varphi \smile \theta) \\ &=(f \times g)^*\mbox{cls}[(\sigma \lambda_p, \varphi)(\sigma \mu_q, \theta)] \end{align} $
I am unsure how to proceed - specifically with the $(f*g)^*$ term. Now I know that $f:X \to Y$ yields a ring homormophism $f^*:H^n(Y;R) \to H^n(X;R)$ given by $f^* \mbox{cls} \varphi = \mbox{cls} f^\# \varphi$
Hopefully all my notation is here is enough to understand the question - if not, let me know and I will edit and update
Edit: As pointed out in the comments below, the above is wrong and the question is really about the naturality of the cross product. I guess my confusion with the $(f \times g)^*$ term still stands. Specifically I know that if f:X \to X' then we have an induced chain map f_\#:S_*(X) \to S_*(X').
We can then apply the functor $\mbox{Hom}(\quad,G)$ to show that $f^\#$ induces a homomorphism f^*:H^n(X';G) \to H^n(X;G) given by $\mbox{cls} \varphi \mapsto \mbox{cls}(\varphi f_\#)$. But I don't know what the map $(f*g)^*$ is?. Obviously is is a map (f*g)^*:H_n(X' \times Y') \to H^n(X \times Y), but where does it take the equivalence classes?