I am asked to describe $Hom(\mathbb{Z}^2, G)$ as a subset of $G \times G$. I interpret this as describing a relationship between the two (i.e. showing they are isomorphic)
G is finite, not necessarily abelian
My intended approach is to generalize the fact that $Hom(Z,G) \cong G$ (which uses the fact that since 1 generates $\mathbb{Z}$ then we can identify each element of $Hom(\mathbb{Z},G)$ with the element $\varphi(1) \in G$ which gives way to an isomorphism between the two structures.
Now, in my case, I know that (1,1) does not generate $\mathbb{Z} \times \mathbb{Z}$, but perhaps I can use (0,1) and (1,0) as my 'basis' of sorts for $\mathbb{Z} \times \mathbb{Z}$; however, I feel there is something wrong with this logic so I would like to know if this seems like the right approach to take before proceeding.
Another idea I thought of was showing that $Hom(\mathbb{Z}^2,G) \cong Hom(\mathbb{Z},G) \times Hom(\mathbb{Z},G) \cong G \times G$ but I'm not sure about the last congruence
Note that the map is completely determined by the images of $(1,0)$ and $(0,1)$, but these commute in the domain, so their images must commute in $G$. This means such hom-set consists of commuting pairs of elements in $G$.
Your last observation is not correct, because $\mathbb Z\times \mathbb Z$ is not the coproduct of $\mathbb Z$ with itself in groups, so that isomorphism doesn't hold. It holds, however, if $G$ is abelian.