Yuri Delanghe's comment is correct, of course, but I am not sure it satisfies your desire for concreteness. In the case of sets, $(X\times Y)\oplus(X\times Z)$ is the disjoint union of $X\times Y$ and $X\times Z$. A standard way of making two sets disjoint is multiplying the first set with $\{0\}$ and the second with $\{1\}$. We use the same technique for $Y\oplus Z$. Now each element of $(X\times Y)\oplus(X\times Z)$ is of the form $((x,a),i)$ where $a\in Y\cup Z$ and $i\in\{0,1\}$. The canonical morphism maps this to $(x,(a,i))$, which is in $X\times(Y\oplus Z)$.
In the case of abelian groups, $X\times(Y\oplus Z)$ and $(X\times Y)\oplus(X\times Z)$ are not isomorphic, in general. There is a nice morphism from $X\times(Y\oplus Z)$ to $(X\times Y)\oplus(X\times Z)$, though. Namely, since for finitely many factors product and coproduct coinside with the direct sum of abelian groups, the elements of $X\times(Y\oplus Z)$ are of the form $(x,(y,z))$ with $x\in X$, $y\in Y$, and $z\in Z$. Such an element gets mapped to $((x,y),(x,z))$ in $(X\times Y)\oplus(X\times Z)$.