Ok I figured it out, it turned out that (if my proof is correct) the proof is really cute.
Firstly, if $x_{1}\neq 0$, then $\{x_{1}\}$ is clearly independent and by part (a) generates $B_{1}$ and thus a basis of $B_{1}$.
Suppose for some $k\geq 2$, $\{x_{1},\cdots, x_{k}\}$ has nonzero elements $\{x_{k,1},\cdots, x_{k,s}\}$ which form a basis for $B_{k}$, then if $x_{k+1}=0$, then the set $\{x_{1},\cdots, x_{k+1}\}$ has nonzero element $\{x_{k,1},\cdots, x_{k,s}\}$ which is hypothesized to be independent. In particular, they can surely generate $x_{k+1}=0$, and thus they can generate $B_{k+1}$. Thus, they form a basis for $B_{k+1}$.
If $x_{k+1}\neq 0$, then the set $\{x_{1},\cdots, x_{k+1}\}$ has nonzero elements $\{x_{k,1},\cdots, x_{k,s}, x_{k+1}\}$, since $\{x_{k,1},\cdots, x_{k,s}\}$ generates $B_{k}$, by part (a), we see that $\{x_{k,1},\cdots, x_{k,s}, x_{k+1}\}$ generates $B_{k+1}$. Now, observe that $\{x_{k,1},\cdots, x_{k,s}\}$ cannot generate $B_{k+1}$, since they are all elements have coordinates being $0$ at and after the $k^{th}$ coordinate, they can surely generate some element in $B_{k+1}$ (in fact they are elements in $B_{k+1}$), but they will miss some elements with the $k^{th}$ coordinate not being $0$ but with coordinate being $0$ starting from $(k+1)^{th}$. Thus, $\{x_{k,1},\cdots, x_{k,s}, x_{k+1}\}$ is the smallest generating set of $B_{k+1}$ and thus it is a basis of $B_{k+1}$.
The result then follows immediately by an induction on $k$.
I am afraid that this answer might not be what you are looking for, because I am thinking algebraically rather than topologically.
In your example, the subgroup is the commutator subgroup $[F,F]$. The vertices of the Schreier graph can be labelled by a transversal to the subgroup, and the obvious choice is $\{a^ib^j: i,j \in {\mathbb Z}\}$. You have edges labelled $a$ and $b$ from $a^ib^j$ to $a^{i+1}b^j$ and $a^ib^{j+1}$, respectively.
The associated free subgroup generators (often called the Schreier generators) are $\{ a^ib^jab^{-j}a^{-i-1} : i,j \in {\mathbb Z}, j \ne 0 \}$.
In a general problem of this type, the vertices of the Schreier graph correspond to the elements of the quotient group $F/\langle S^F \rangle$ of $F$, which is the group defined by the presentation $\langle a,b \mid S \rangle$.
Best Answer
Most proofs of the Nielsen-Schreier theorem proceed by following a recipe for finding generators of the subgroup $H$ in $F$ and then proving that they are free generators.
You start by finding a (right) Schreier (= prefix closed) transversal of $H$ in $F$. Let $X$ be a set of free generators of $F$ and, for $g \in F$, let $\bar{g}$ denote the unique element of $U$ with $Hg = H\bar{g}$.
Then the subset of non-identity elements of the set $$\{ux \overline{ux}^{-1} : u \in U, x \in X \}$$ freely generates $H$.
In your example, we can take $U = \{1,a\}$, leading to the free generating set $\{b,a^2,aba^{-1}\}$ of $H$, which is the same as that found in the answer by kabenyuk.
You can check normality of a subgroup $H$ of $F$ by constructing the permutation representation of $F$ on the cosets of $H$ in $F$, and then checking whether the generators of $H$ all act trivially on this set.
The above method of testing for normality is fine for subgroups of moderately small finite index, but for arbitrary finitely generated subgroups, you can use the fact that membership testing in such subgroups is possible (using methods based on Stallings' Folding) and then testing membership of $xgx^{-1}$ and $x^{-1}gx$ in $H$ for all $x \in X$ and generators $g$ of $H$.