If I have a free abelian group G with basis $\{g_i\}$ and H a subgroup of G. Then H is free abelian and a basis is given by the elements of $\{g_i\}$ that are in $H$, right?
Then under what condition is $G/H$ free? Is a basis given by $\{g_i+H\}$, for all the $g_i$ that are not in H?
On
No element of the basis needs to belong to $H$:
Consider the free abelian group $G =\mathbb{Z}^{\oplus2}$ with the standard basis $(1,0),(0,1)$ and the subgroup $H \subset G$ given by the diagonal. Then $H \cong \mathbb{Z}$, but none of the standard basis vectors are in $H$.
For a free abelian group $G$ of finite rank $n$ you can at least prove the following:
Let $H \subset G$ be a subgroup of say rank $m$. Then there exists a basis $g_1,\dots,g_n$ of $G$ and integers $d_1 \mid \dots \mid d_n$ such that $d_1g_1,\dots,d_mg_m$ is a basis for $H$.
One can show that the quotient is free if you can extend a chosen basis of $H$ to a basis of $G$.
Your first claim is wrong. Take the free abelian group $\mathbb{Z}$, which has basis $\left\lbrace 1\right\rbrace$. Then a subgroup is of the form $n\mathbb{Z}$ which is free, generated by $\left\lbrace n\right\rbrace$ which is not a subset of the generators of $\mathbb{Z}$.
You see that in this example no quotient is free. In fact, the quotient $G/H$ is free precisely if and only we can extend a basis of $H$ to a basis of $G$.