For regular languages A and B, determine whether B might match early in (A B)
If I translate your description properly, here is what you want to ask:


  Given two regular language $A$, $B$, is there a prefix of a word $b$ in $B$ that is a suffix of a word in $A$ so that the rest of $b$ is a prefix of another word in $B$?


Formally:


  Given regular $A$ and $B$, are there $a = wx \in A$ and $b = yz \in B$ so that $xy \in B$?


Deciding this boils down to deciding whether

$\qquad B \cap (\operatorname{suff}(A) \cdot \operatorname{pref}(B) ) = \emptyset$.

Now we employ some closure properties: $\mathrm{REG}$ is closed against right and left quotient, concatenation and intersection, so with

$\qquad \operatorname{suff}(A) = A \backslash \Sigma^*$ and 
$\operatorname{pref}(B) = B / \Sigma^*$

the set we want to check for emptiness is regular, and therefore the check is indeed decidable.

In order to actually do this, I suggest you build automata for $\operatorname{suff}(A)$ and $\operatorname{pref}(B)$, concatenate them, intersect the result with the automaton for $B$ and use the standard check for emptiness (is a final state reachable from the initial state?).

Note that you never need DFA, all of this works nicely with NFA. Therefore, no exponential state explosion occurs; intersection multiplies the state numbers, though.