The set of all functions $f : \Bbb{N} \to \Bbb{Z}$ such that $f(n) = 0$ for all $n \geq N_f$ for some $N_f \in \Bbb{N}$ dependent upon $f$, forms a $\Bbb{Z}$-module.
Proof. Let $M$ be the set of all such functions. Then clearly, $0 \in M$. Suppose that $f, g \in M$
$$ (f-g)(n) = 0, \ \forall \ n \geq \text{max}(N_f, N_g) $$
Done since scalar multiplication clearly holds, and we've shown closure of a subset $M \subset \Bbb{Z}$-module $\{ \text{functions } h : \Bbb{N} \to \Bbb{Z}\}$ under subtraction.
What is a basis for $M$?
Let $f_n:\mathbb{N}\to \mathbb{Z}$ the function that assigns to each $k$ the value
$f(k):= \delta_{kn}$
Then $\{f_n: n\in \mathbb{N}\}$ is a base for your $\mathbb{Z}-$module $M$.
In fact if you consider $f\in M$, then
$f=\sum_{k}^{N_f}f(k)f_k$