Two way finite state Automaton NP Compleness definition clarification? INSTANCE: A two-way nondeterministic finite state automaton A over single Alphabet I.
QUESTION: Is there an string x (over I) such that A accepts x ? 

Since a 2-way NFA with N states, there’s an equivalent nfa with n^(n+1) states. So that’s we get the exponential time from.

I am clear about that, but if I understand correctly the size of the string accepted by the 2Way NFA would still be (in the worst case) polynomial to number of states in the 2Way NFA. Cause if the smallest string accepted was exponential in size, the verification will also take exponential time and thus the problem by definition would not be NP Complete ? Or is there a mistake in this reasoning?
For a problem to be in NP, there needs to be a polynomial time verifier for it. This polynomial time verifier need not be the "obvious" one. Ruling out one specific verifier doesn't prove that the problem is not in NP.

For a problem to be NP-complete, it needs to satisfy two properties: (i) it needs to be in NP, and (ii) it need to be NP-hard. You can look these concepts up in many sources. In particular, the fact that there is some natural exponential time algorithm for a problem doesn't imply that it is NP-complete, though it is true that every problem in NP has an exponential time algorithm.

Unfortunately I am not versed well enough in two-way automata to answer your specific question, but my comments above imply that the answer doesn't necessarily have a direct bearing on whether the emptiness problem for 2NFAs is NP-complete or not (or even in NP or not).