Maybe the clearest way to see that $(\mathbb Z \oplus \mathbb Z) / \langle (2, 2) \rangle = \mathbb Z \oplus \mathbb 2 \mathbb Z$ is to perform a change of basis on $\mathbb Z \oplus \mathbb Z$.
Usually when we think of $\mathbb Z \oplus \mathbb Z$, we think in terms of the standard basis, $\{(1, 0), (0, 1)\}$. But there are other bases for $\mathbb Z \oplus \mathbb Z$!
One such alternative basis is $\mathcal B = \{(1, 1), (0, 1) \}$. Now $(2, 2)$ can be expanded in terms of this alternative basis as $(2, 2) = 2(1,1) + 0(0, 1)$. In other words $(2,2) = (2, 0)_{\mathcal B}$ (if that notation makes sense).
Thus $(\mathbb Z \oplus \mathbb Z) / \langle (2, 2) \rangle$ is isomorphic to $(\mathbb Z \oplus \mathbb Z) / \langle (2, 0) \rangle$ (via the above change of basis), and obviously $(\mathbb Z \oplus \mathbb Z) / \langle ( 2, 0) \rangle$ is isomorphic to $\mathbb Z_2 \oplus \mathbb Z$.
(By the way, there is a typo in your post - it's not isomorphic to $\mathbb Z \oplus 2 \mathbb Z$!)
This begs the question:
Given a subgroup $G$ of $\mathbb Z^{\oplus n}$ (defined by a set of generators), is there a systematic algorithm that will find an alternative basis $\mathcal B = \{ \mathbf v_1, \dots, \mathbf v_n \}$ for $\mathbb Z^{\oplus n}$, a $k \in \{0, \dots, n\}$ and integers $c_1, \dots, c_k$ such that $\{ c_1 \mathbf v_1, \dots, c_k \mathbf v_k \}$ is a basis for $G$? Because if so, then $\mathbb Z^{\oplus n} / G \cong \mathbb Z_{c_1} \oplus \dots \oplus \mathbb Z_{c_k} \oplus \mathbb Z^{n - k}$.
[In the above example, $n = 2$, $G = \langle (2, 2) \rangle$, $\mathcal B = \{ (1, 1), (1,0) \}$, $k = 1$ and $c_1 = 2$.]
Yes, such an algorithm exists, and it's called the Smith normal form algorithm. Take a look at the final few pages of these notes for an explanation.
