Why is $\mathbb{Z}_2 \ast \mathbb{Z}_2$ not a free group? I recently start reading Hatcher's book for self-study. On page $46$ it gives such an example that is a free product and not a free group. 
I don't quite understand the explanation given in the book. Should I show that any subset of $\mathbb{Z}_2\ast\mathbb{Z}_2$ cannot be the basis for a free group? 
Secondly, what is the relationship between free abelian groups and free groups?
 A: Exercise: There is a nonidentity element in your group with a finite conjugacy class. Nonabelian free groups have no such element!
A: A free group has only generators, and no relations. A free product of two groups means something rather different: in a sense, if $A$ and $B$ are groups, the free product of $A$ and $B$, written $A*B$ is as close to being free as it can be, given that $A$ and $B$ are subgroups of it: that is, there are no extra relations in the free product part from those already given on $A$ and $B$ separately. Hence the free product $A*B$ is itself a free group if and only if $A$ and $B$ are already free groups. The free product of two cyclic subgroup of order $2$ therefore has no chance to be a free group ( it is in fact easily checked to be an infinite dihedral group). Another common group theoretic construction which involves the word "free" is the free product with amalgamation, or just amalgam, of two groups. A good place to read about these things is J.P. Serre's book "Trees".
A: (Your first question was answered by i.m. soloveichik in the comments.)
A free group generated by a set of generators is, in a sense, the largest group you can make out of those generators.
A free abelian group generated by a set of generators is, likewise, the largest abelian group you can make with those generators. 
Being "free" essentially means the group is not subject to any relations. Elements having finite order is a special case of satisfying the relation $a^n=1$ (or $na=0$, when our group is written additively). 
The nonabelian one will usually be larger, because they are not subject to the simplification that commutative operations provide. 
I'm not an abelian group person, but I guess that if you take the free group on some generators and mod out the normal subgroup generated by elements of the form $aba^{-1}b^{-1}$, then the quotient group will be the same as the free abelian group on those generators. (Modding out that subgroup forces the quotient to be commutative.)
A: This only concers your last question about free and free abelian groups.
Here is a way to think about this.
Consider the forgetful functors 
$$Groups \to Sets \text{ and } Abelian Groups \to Sets$$
that just view the groups as sets and group homomorphisms as maps of sets.
Now I claim that these functors have left adjoints, called the free functors.
Those take a set and give you a group or an abelian group (depending on if you view an abelian group to be an object of Groups or abelian groups).
It is in general always a good intuition that a "free" object should come from a left adjoint to a forgetful functor. I.e. if you have some object with extra structure then you can always forget this extra structure and just remember "the rest" of the object. A free object in your category is then just something that comes from the left adjoint of your forgetful functor (this left adjoint does not always have to exist, though).
Note that this not only depends on your starting category, but also on "how much" you forget. For example when you have a an abelian group you can first forget that it is abelian and just view it as a group, and then forget the group structure, or you could directly forget everything. 
My point being, both free and free abelian groups are (in the previous sense) just free objects in the category of abelian groups respectively of groups, freeness referring to the forgetful functor to sets.
