In your example, take $\;B:=\{\,(2,3)\,,\,(1,2)\,\}\;$ , then show:
1) $\;B\;$ is a $\,\Bbb Z\,-$ linearly independent set and is thus a basis of $\;\Bbb Z\times\Bbb Z\;$ ;
2) $\;C:=\left\{\,1\cdot(2,3)\,,\,6\cdot(1,2)\,\}=\{\,(2,3)\,,\,\,(6,12)\,\right\}\;$ are $\;\Bbb Z\,-$ linearly independent and a basis of $\;2\Bbb Z\times3\Bbb Z\;$ as required.
One way to think about the free abelian group on a set $X$ is that you are taking the elements of $X$ as a starting point, and then making a group operation on those elements that is abelian, but where the elements don't satisfy any relations other than the ones that they absolutely have to to make an abelian group. In particular, the elements of $X$ are viewed here as something like "atoms", and we're building a totally new group operation on them. It doesn't matter at all what they are, just how many of them there are.
In your second example the fact that we built a group from the numbers $1, 2, 3$ doesn't mean that the group operation we're building has anything to do with the usual arithmetic on those numbers. It might help to give the new group operation a new name, say $\ast$. Then elements of the free abelian group on $\{1, 2, 3\}$ will have the form $(a1)\ast(b2)\ast(c3)$, with $a, b, c \in \mathbb{Z}$, and with no further simplification possible. The group we get this way is isomorphic to $\mathbb{Z}^3$, via the function that sends $(a1)\ast(b2)\ast(c3)$ to $(a, b, c)$.
Best Answer
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 $G$.