Definition of Relative Version of Cross Product in Homology On pg. 210 of the hard copy, and on pg 218 on the online copy of Hatcher's Algebraic Topology is defined the notion of relative cross product as follows:

Definition. Let $(X, A)$ and $(Y, B)$ be pairs of topological spaces, and $p_1:X\times Y\to X$ and $p_2:X\times Y\to Y$ be the projection maps. Define $\times :H^k(X, A; R)\times H^l(Y, B; R) \to H^{k+l}(X\times Y, A\times Y\cup X\times B; R)$ as
  $$(a, b)\mapsto p_1^*(a)\cup p_2^*(b)$$
  where '$\cup$' denotes the cup product.

I do not understand how this definition makes sense. The map $p_1$ does not give us a map of pairs $(X\times Y, A\times Y\cup X\times B)\to (X, A)$, since the image of $A\times Y\cup X\times B$ under $p_1$ is all of $X$, and not $A$. Similarly for $p_2$.
So it is not clear what is meant by $p_1^*$ and $p_2^*$ here. Can somebody please clarify what is Hatcher trying to say here.
Also, what makes sense is using $p_1:(X\times Y, A\times B)\to (X, A)$ and $p_2:(X\times Y, A\times B)\to (Y, B)$ to get
$$\times : H^k(X, A; R)\times H^l(Y, B; R)\to H^{k+l}(X\times Y, A\times B; R)$$
by writing $$(a, b)\mapsto p_1^*(a)\cup p_2^*(b)$$.
Is there a good reason why the second definition cannot be found in literature?
 A: If you have a tripled ($X, A, B)$ with $A$ and $B$ open in $X$ you can define a relative cup product 
$$H^*(X, A) \times H^*(X, B) \to H^*(X, A \cup B)$$
Define this on (co)chain level by taking cochains $\alpha, \beta$ from $C^k(X, A)$ and $C^\ell(X, B)$ respectively, cup product of which lives in the subgroup $C^{k+\ell}(X, A + B) \subset C^{k+\ell}(X)$ of cochains vanishing on sums of chains in $A$ and chains in $B$. $\alpha \cup \beta$ does indeed do so (this is just checking the definitions). To push $\alpha \cup \beta$ into something co homologous to a cochain living in the subgroup $C^{k+\ell}(X, A \cup B)$ is all what remains. This can be done by noting that the inclusion $C^*(X, A \cup B) \hookrightarrow C^*(X, A + B)$ is an isomorphism on homology (five lemma + (co)chain-level statement of excision). 
The geometric point is that if you think of $\alpha \cup \beta$ as a piecewise-linear function on $X$ (appropriately triangulated), it vanishes on sums of simplices in $A$ and simplices in $B$. It vanishes on any chain on $A \cup B$ (which need not be like that: the chain could contain simplices which goes half into $A$ and half into $B$) because that's what excision says: if you triangulate the chain, refine it enough, it'd consist entirely of simplices of $A$ and simplices of $B$, on which $\alpha \cup \beta$ vanishes. That's all what "upto cohomology" happens.

That being said, $p_1^*(a)$ lives in $H^*(X \times Y, A \times Y)$ and $p_2^*(b)$ lives in $H^*(X \times Y, X \times B)$. Their cup product lives in $H^*(X \times Y, A \times Y \cup X \times B)$.
