Let's start with a simpler case: $\mathbb{Z}/2\mathbb{Z}$. Here, $2\mathbb{Z}=\{\ldots,-4,-2,0,2,4,\ldots\}$. So, the cosets of $2\mathbb{Z}$ in $\mathbb{Z}$ are the $z+2\mathbb{Z}=\{\ldots,z-4,z-2,z,z+2,z+4,\ldots\}$. If that $z$ would happen to be $2$, we'd get back $2\mathbb{Z}$ (just shifted over one). Thus $2+2\mathbb{Z}=2\mathbb{Z}$. Since you can write any number as $2m+n$ where $m$ is some integer and $n$ is $0$ or $1$, we can characterize all cosets of $\mathbb{Z}/2\mathbb{Z}$ by $2m+n+2\mathbb{Z}=n+2m+2\mathbb{Z}=n+2\mathbb{Z}$, so the only two cosets are $2\mathbb{Z}$ and $1+2\mathbb{Z}$.
Now, keeping that process in mind, let's do $(\mathbb{Z}\times\mathbb{Z})/\langle(2,2)\rangle$. We have $\langle (2, 2 ) \rangle =\{\ldots,(-4,-4),(-2,-2),(0,0),(2,2),(4,4),\ldots\}$ Now if somebody adds $(1,0)$ to each of these numbers, we get $(1,0)+\langle (2, 2 ) \rangle =\{\ldots,(-3,-4),(-1,-2),(1,0),(3,2),(5,4),\ldots\}$ If we keep adding them, we will keep shifting the left and right numbers apart: they'll never equal $\langle (2, 2 ) \rangle$ again because nothing modifies the second coordinate. Thus, the group is infinite.
Now, is it cyclic? It's infinite, any any infinite cyclic group is isomorphic to $\mathbb{Z}$. This is where the part where you can add $(1,1)+\langle(2, 2 )\rangle$. You can easily verify this has order $2$ in just the same way as the $2\mathbb{Z}$ example, and $\mathbb{Z}$ contains no elements of finite order, so it must not be cyclic.