Intersection Non-Emptiness for DFA's is as follows:
Input: A finite list of DFA's $D_1$, $D_2$, ..., $D_k$.
Question: Does there exist a string $w$ such that for every $i \in [k]$, $D_i$ accepts $w$? In other words, is the intersection of their associated regular languages non-empty?
Intersection Non-Emptiness is a classic PSPACE-complete problem (Kozen 1977 - "Lower bounds for natural proof systems")
Relevance: There is a nice and simple parameterized reduction from intersection non-emptiness for one-way DFA's to non-emptiness for two-way DFA's.
Pick the number of DFA's to be the parameter for Intersection Non-Emptiness and the number of turns (switches from moving left to right or right to left) as the parameter for Non-Emptiness for two-way DFA's.
Claim: Intersection Non-Emptiness for $k$ DFA's is reducible to Non-Emptiness for $(2k-2)$-turn two-way DFA's. (I believe that there is a related reduction for the other direction too.)
Given DFA's $D_1$, $D_2$, ..., $D_k$, we can construct a $(2k-2)$-turn two-way DFA that evaluates each of the DFA's on the input string one at a time.
First, it evaluates $D_1$ on the input. Then, it moves the tape head back to the beginning and evaluates $D_2$ on the input. Then, it moves the tape head back to the beginning and evaluates $D_3$ on the input. ... Finally, it moves the tape head back to the beginning and evaluates $D_k$ on the input.
If all of them accept, then it does the evaluation on all of them and then accepts. If one of them rejects, then it stops (doesn't finish evaluating on all of them) and immediately rejects.
Hardness: If you can solve Intersection Non-Emptiness for $k$ DFA's in less than $n^k$ time, then the strong exponential time hypothesis is false.
Related link: https://cstheory.stackexchange.com/questions/29142/deciding-emptiness-of-intersection-of-regular-languages-in-subquadratic-time/29166#29166
Therefore, by the reduction, if you can solve non-emptiness for $(2k-2)$-turn two-way DFA's in less than $n^k$ time, then the strong exponential time hypothesis is false as well.
Conclusion: If you were to find a faster algorithm for non-emptiness for two-way DFA's, then that would lead to a more efficient simulation of non-deterministic machines. Let me know if you have any thoughts to share. Thank you for asking the question! :)