Let $F_2=\langle a,b\rangle$ be the free group generated by $a$ and $b$, $W^*$ the words of $\{a,\overline{a},b,\overline{b}\}$ and $w \in W^*$ be a reduced word (I am actually looking for an "asymmetric word" so you can assume $w=a^k b$ for some large enough $k$ if this helps). The question is then:
Can we find two generators $s,t \in F_2$ satisfying the following conditions:
- $s$ and $t$ are cyclically reduced
- $s$ and $t$ have as common start $w$, i.e. we have as reduced words $s=w s'$ and $t=w t'$ with $s',t' \in W^*$ and $s'$ and $t'$ are non-empty and start with different letters
- $s$ and $t$ end with different letters
This question arose in the construction of a counterexample, namely I need in the Cayley-graph of the free group generators whose translation-axes intersect in a most asymmetric way (or are far away, which is not possible for $F_2$). After several (obviously fruitless) attempts to construct such $s$ and $t$ I forward this question now to you.
If by "two generators" you mean "a generating pair", so $\langle s,t\rangle=F(a, b)$, this is impossible.
The (folded) Stallings' graph corresponding to your a pair $(s, t)$ is a $\theta$-graph, with the central line corresponding to $W$ and the two other edges corresponding to $s′$ and $t′$. The assumptions in the question mean that all three edges have length $\geq1$. However, bases of $F(a, b)$ have Stallings' graph equal to the bouquet with a single vertex and two loop edges, labelled $a$ and $b$ respectively, and so $\langle s,t\rangle\neq F(a, b)$ as claimed.