DFA intersection algorithm for special cases
Yes, there are some cases of the DFA non emptiness insersection problem that are inside P. My master's thesis is devoted to this question, but unfortunately it is in French. However, most of the results have appeared here in $[2]$.

When the alphabet is unary, then the problem is L-complete when each DFA has at most two final states, and NP-complete otherwise. Most of the other cases are restriction on the transition monoids of the automata. For example, for abelian group transition monoids, the problem is in $\text{NC}^3$ when each DFA has at most one final state, and NP-complete otherwise; for elementary 2-group transition monoids, the problem is $\oplus$L-complete when each DFA has at most two final states, and NP-complete otherwise.



Let me now address your more precise question, which can only be found in $[1]$. Suppose you are given DFAs working over $\{0,1\}$ and shaped as trees, i.e there exists a state $u$ (initial state) such that for each state $v$ there exists a unique path from $u$ to $v$. Then, deciding the intersection non emptiness is:


L-complete for one final state in each DFA,
NL-complete for two final states in each DFA, and
NP-complete for three or more final states in each DFA.


The hardness results still hold even if you "fork" respectively 0, 1 or 2 times (this is your $K$). Now if your DFAs are directed acyclic graphs instead of trees, then the problem is NP-complete even with one final state in each DFA and $K=2$; the reduction is quite straightforward and is from Monotone 1-in-3 3-SAT.

Therefore, no, I don't think there is an efficient algorithm for your problem.

Now, if the number of automata is fixed, you might want to discuss with Michael Wehar who has recently published $[3]$.



EDIT: Since OP edited his question, let me clarify my answer with his new requirements. Consider the NP-complete problem Monotone 1-in-3 3-SAT where you are given a formula in 3-CNF without negation, and where you have to determine whether there is an assignment that makes exactly one variable true in each clause. You can reduce this problem to the non emptiness intersection problem as follows. For example, for the clause $x_2 \lor x_3 \lor x_5$, you build the following automaton:

$\hskip2in$

Note that the automata are trees (and hence DAGs), are levelled, and have three final states. Actually, the three final states coud be merged into a single one, if one is satisfied with DAGs. Moreover, only two states have two (distinct) outgoing transitions.


Michael Blondin. Complexité raffinée du problème d'intersection d'automates, M.Sc. thesis, Université de Montréal, 2012.
Michael Blondin, Andreas Krebs et Pierre McKenzie. The Complexity of Intersecting Finite Automata Having Few Final States, Computational Complexity (CC), 2014.
Michael Wehar. Hardness Results for Intersection Non-Emptiness. ICALP, 2014.