Every abelian group G is the homomorphic image of a free abelian group of rank $|X|$, where $X$ is a set of generators of $G$.
Here is the proof that Hungerford provides in his Algebra book:
Let $F$ be the free abelian group on the set $X$. Then $F = \sum_{x \in X} \Bbb{Z}x$ and $rank~~ F = |X|$. By theorem 1.1 the inclusion map $X \to G$ induces a homomorphism $\overline{f} : F \to G$ such that $1x \mapsto x \in G$, whence $X \subseteq Im(\overline{f})$. Since $X$ generates $G$ we must have $Im(\overline{f})=G$.
Here is theorem 1.1.
So, in the present case, we have that (ii) is true, so (iv) must be true. I.e., there exists some map $i : X \to F$ satisfying the property mentioned in (iv). Taking $f: X \to G$ to be the inclusion map, there must exist a unique homomorphism $\overline{f} : F \to G$ such that $\overline{f} i = f$. My first question is, how do we get from this that $\overline{f}(1x) = x$, seeing as we have no idea what the map $i$ does other than satisfy $\overline{f}i = f$ (note: I tried to use this to prove the $\overline{f}(1x)=x$, but I had no luck).
My second question is, how does $Im(\overline{f})=G$ prove that $G$ is the homomorphic image $F$? The equation $Im(\overline{f})=G$ doesn't even seem to make sense since $Im(\overline{f})$ is a subgroup of $F$.
For your first question : when $F=\sum_{x\in X} \Bbb Z$, the map $i:x\to F$ is defined by $x\mapsto 1x$ (this should probably appear in the proof of Theorem 1.1 in the book, in the $(iii)\Rightarrow (iv)$ part). So for every $x\in X$ you have $$\overline{f}(1x)=(\overline{f}i)(x)=f(x)=x.$$
For your second question : $Im(\overline{f})$ is the image of $\overline{f}$, i.e. the set $\{\overline{f}\left(\sum_{x\in X}\alpha_x x\right)\mid y\in F\}$. Since $\overline{f}$ is a group homomorphism $F\to G$, this is a subgroup of $G$, and since it contains $X$, it must in fact be equal to $G$. This is exactly what "being the homomorphic image of a group" means.