Suppose $\xi \in H^{p_2}(X),\eta \in H^{q_2}(Y),x \in H_{p_1+p_2}(X),y \in H_{q_1+q_2}(Y)$. Then $$(\xi \times \eta) \frown ( x \times y) = (-1)^{p_2 q_1}(\xi \frown x) \times (\eta \frown y)$$
Cross product is defined by Eilenberg-Zilber homotopy equivalence. That is \begin{align} S^{p}(X) + S^{q}(Y) &\overset{\otimes}\rightarrow S^{p}(X) \otimes S^q(Y) \overset{Eilenberg-Zilber}{\longrightarrow} S^{p+q}(X \times Y)\\ (x,y) &\mapsto x\otimes y \mapsto EZ(x\otimes y) \end{align} (So as in $S_*(X)$)
And cap product is given by \begin{align} S^q(X) \times S_{p+q} \overset{\frown}\longrightarrow S_p\\ \xi \frown \sigma = \sideset{_p}{}{\sigma} \langle\xi,\sigma_q\rangle \end{align} where $\langle\cdot,\cdot\rangle$ is Kronecker product.
We can make EZ homtopy equivalence concrete by construct \begin{align} S^{p+q}(X \times Y) &\overset{Eilenberg-Zilber}{\longrightarrow} S^{p}(X) \otimes S^q(Y)\\ \sigma &\mapsto \sum_{i+j = p+q}\rho_1(\sideset{_i}{}{\sigma}) \otimes \rho_2(\sigma_j) \end{align}.
Therefore $$(\xi \times \eta) \frown ( x \times y) = \sideset{_{p_1+q_1}}{}{(EZ(x\otimes y))} \langle EZ^{\sharp}(\xi\otimes \eta) ,(EZ(x\otimes y))_{p_2+q_2} \rangle$$
The difficulty is $(EZ(x\otimes y))_{p_2+q_2}$ which is obtained by truncating last $p_2+q_2$ dimensional face. This operation is not a chain map.