I am trying to prove whether or not the ideal generated by $\langle (2,2)\rangle$ is a prime ideal of $\mathbb Z_4\times \mathbb Z_4$?
My issue is I'm not sure how to do the coordinate multiplication: would the ideal look like $\{(0,0),(2,2)\}$ or would it be $\{(0,0), (2,2), (2,0), (0,2) \}$? I think that either way it is a prime ideal, but I want to make sure that I am doing the multiplication for the ideal correctly.
Given an element $a$ in a ring $R$ the ideal $\langle a\rangle=\{ra\ |\ r\in R\}$
So in your case you just have to take each element of $\mathbb Z_4 \times \mathbb Z_4$ and multiply it with $(2,2)$ which will give you the ideal that you have written second.
What you wrote first is the cyclic subgroup of $\mathbb Z_4\times \mathbb Z_4$ generated by $(2,2)$
Also note that the given ideal is not prime for the following reason - we know that an ideal $P$ is prime if whenever $fg\in P\implies f\in P$ or $g\in P$. However your given ideal contains the product $(3,0)\cdot (0,3)=(0,0)$ but neither $(0,3)$ nor $(3,0)$ is in it.