Let $M$ be an $R$-module. Then $M$ is cyclic $R$-module if and only if there is a left ideal $A$ in $R$ such that $M$ isomorphic to $R/A$.
I am not sure why this is true. One proof of this says that $R/A = \langle 1+A \rangle$, but I am not sure why this would be true; indeed, given the addition in $R/A$, would the generating not consist only of integer multiples of $1$? Am I not understanding this correctly?
The element $1 + A$ generates $R/A$ as an $R$-module. Indeed, for any $r + A \in R/A$, we have $$ r + A = (r\cdot 1) + A = r(1 + A), $$ by how scalar multiplication on the quotient module is defined. Thus, $R/A$ is clearly a cyclic module.
The integers are not involved here; this is not a module over $\mathbb{Z}$, but rather over a general ring $R$, which may be causing the confusion.
Now, note that this proves the backwards direction; if $M$ is isomorphic to $R/A$, then it is cyclic, since $R/A$ is cyclic. For the forwards direction, suppose $M$ is cyclic, generated by $x \in M$ (so $M = \langle x \rangle$). Consider the map $x \mapsto 1_R + A$, where $A$ is a certain ideal (for this, see https://en.wikipedia.org/wiki/Cyclic_module).