Is this group a free group on these generators? Let $F_2$ be the free group on two elements $a,b$ and $C_2=\{0,1\},$ the cyclic group of order $2.$
Let $H$ be the kernel of the unique homomorphism $\phi:F_2\to C_2^2$ with $\phi(a)=(1,0)$ and $\phi(b)=(0,1).$
Any subgroup of a free group is free, by this theorem.
In this answer I prove that $H$ is generated by: $$a^2,b^2,(ab)^2,(ba)^2, (a^2b)^2, (ab^2)^2$$
Is $H$ free on these generators?
This is equivalent to asking if the group is free on:
$$a^2,b^2,(ab)^2,(ba)^2,ab^2a,ba^2b,$$
which might be easier, because these are all “words” of length at most $4.$
But I have no idea how to show there is no reduce the set of generators.
 A: No. The first five of your six  generators are free generators of $H$. Note that
$$ba^2b = (ba)^2 (ab^2a)^{-1} (ab)^2.$$
In fact by the Schreier formula, the free rank of a subgroup of index $r$ in $F_n$ is $n(r-1)+1$, which gives $5$ for the free rank of $H$.
A: There's a very elegant algorithm that can, among other things, find a minimal generating set for any finitely generated subgroup of a free group. It's described in the following paper:
Stallings, John R. Topology of finite graphs. Invent. Math. 71 (1983), no. 3, 551–565.
The paper is absolutely worth a read - easily one of my favorites - but the algorithm can be described rather compactly. Consider a directed graph with an identified vertex (in the middle) and edges labelled by generators of a free group, like this:

We can associate any path $P$ along such a graph with an element $e_P$ of a free group - where, at each edge of the path, we multiply by the generator associated to that edge if we traverse it forwards and by the inverse of that generator if we go backwards. For instance, the path traversing the righthand loop clockwise is associated to $aba$. If we consider the set of all loops based at the marked point, the elements associated to these loops will form a subgroup - since we can multiply any two loops together by concatenation and invert an element by traversing the loop backwards. So here, I've depicted the group $\langle a^2, aba\rangle$ for simplicity.
Note that it's trivial to draw a graph corresponding to any finitely generated subgroup - just draw a bouquet of loops meeting at the marked point, with each loop marked and directed in such a way that traversing the loop would multiply by a generator of the subgroup.
The clever step is to notice that, if we have any vertex with two outgoing (or incoming) edges with the same label, we can identify the vertices to which those edges connect without changing the associated group. This would let us merge the top two vertices in the graph, for instance:

And we could repeat the same, noting that the marked vertex has two incoming edges labelled $a$, allowing us to identify those too:

At this point, we have simplified as much as we can. Of note is that this is now a transition graph for a finite state machine that recognizes words in the subgroup. Having done this reduction, finding a minimal generating set is easy - take any spanning tree $T$ in the graph and then, for each edge $e$ not in $T$, consider the paths $p$ and $q$ connecting the base point to the source and sink of $e$ respectively. The concatenations $peq$ are loops based at the marked point and the elements associated to these loops are a minimal generating set for the subgroup.
In this case, it's not too interesting - if we choose the spanning tree to consist of the lower path between the nodes, we get $\langle a^2, aba^{-1}\rangle$ as the generating set, and if we choose the upper path, we'd get $\langle a^{2}, a^{-1}ba\rangle$, which isn't shocking at all. The point, however, is that this algorithm can be applied to absolutely any subgroup of a free group, and gives an easy proof of various results about free groups.
If you applied this algorithm to the group in question, you'd get the following after all is said and done - where we'll consider the bottom-left point to be the base point (which I forgot to mark):

You might recognize this as the Cayley graph of $C_2^2$ - which is, of course, not a coincidence at all, since to decide if an element is in a normal subgroup, you can just carry out multiplications in the quotient group and see whether you end up at the identity, which is what this graph suggests. You would get a similar graph on cosets if the group had finite index, even if it weren't normal.
We can directly extract a minimal set of generators from the Cayley graph: choose a spanning tree - for instance, we could take the edges corresponding to the paths $ab$ and the paths $b^{-1}$ based at the base points. Then, for each other edge, we form a generator. This particular choice of spanning tree would yield the following generating set:
$$\langle a^2, ab^2a^{-1}, (ab)^2, aba^{-1}b, b^2\rangle$$
Which only has five elements.
Note that if you wanted to recover how to write an arbitrary element of the subgroup in terms of these generators, you would find the loop corresponding to that arbitrary element (which is easy). Each edge of this loop outside of the spanning tree corresponds to one of the elements of the previous generating set (or its inverse, if you went backwards over the edge) - and if you multiply together those corresponding generators, you'll get the element you started with.
Also note that this gives an easy way to prove the Schrier formula mentioned in other answers (...though a little more work needs to be done to make all of this rigorous) - and it does so without resorting to algebraic topology (even if the connection to algebraic topology is fairly clear).
A: By the Nielsen-Schreier theorem/formula you quote, the subgroup $H$ has index $e = 4$ in the free group $F_{2}$ of rank $n = 2$, and thus is free of rank $1+e(n-1) = 5$.
