Confusion related to definitions involving free groups From Wikipedia

...the free group $F_{S}$ over a given set $S$ consists of all expressions (a.k.a. words, or terms) that can be built from members of $S$, considering two expressions different unless their equality follows from the group axioms (e.g. $st = suu^{−1}t$, but $s ≠ t$ for $s,t,u \in S$). The members of $S$ are called generators of $F_{S}$.

I don't understand  the distinction between " $a$ and $b$ freely generate a group" , "$a$ and $b$ generate a free group" and just checking if a group is free. For example I thought that if $a$ and $b$ freely generate a group then that group is free and vice versa. However I have seen statements like " $a$ and $b$ freely generate a free subgroup of..." quite a few times and so there seems to be some distinction between the terms. If so could you please provide an example where one of the conditions hold bit the other doesn't?
If, for example, $S=\{a,b\}$, but $a^{3}=1$ and $b^{2}=1$ is it the case that $S$ cannot generate a free group even if any word written using only the letters $a$,$a^2$ and $b$ (excluding powers of these) has a unique representation? 
If we say that we are considering words in $\{a,b\}$ does that mean that we consider words using $a$,$b$, $a^{-1}$ and $b^{-1}$ as letters and identifying $b^2=1$ and $a^3=1$ in the process of reducing the letter or that we use any integer power of and b and only reduce identities that hold by group axioms independent of the actual group structure we are considering?
I apologize if the question is confusing but I am quite confused myself. If you think there is any way of clarifying the question please feel free to suggest it.
Thank you.
 A: As I understand it, the statement that "a set $S$ freely generates the group $G$" means that the group generated by $S$ is free: in other words, there are no relations between elements of $S$. In your example, $a$ and $b$ would not freely generate the group because they satisfy the relations $a^3=1$ and $b^2=1$. (When I say "relation", I exclude the obvious group axioms such as $aa^{-1}=1$. But excluding these follows from the definition of reducing words in a free group.)
It is very much possible for a subgroup of a nonfree group to be free. Consider, for example, the infinite dihedral group $D_\infty$, which has presentation
$$
\langle r,s\mid s^2 = 1, srs=s^{-1}\rangle.
$$
The subgroup of $D_\infty$ generated by $r$ is free and isomorphic to $\mathbb Z$; so $\{a\}$ freely generates the subgroup.
A: Think of the additive group of integers; it is free of rank 1. The integer 1 is its free generator. The set of integers, say 2 and 3, are generators but not free generators. 
Edit: Wikipedia is a great source but cannot replace a textbook. Consider reading Lyndon and Schupp "Combinatorial group theory" or Karras, Magnus, Solitar, same title. This should help. 
