Complexity of #SAT for monotone DNF formulae whose hypergraph is a hypertree
Observe that the negation of a (monotone) CNF-formula F is a (monotone) DNF G which has the same hypergraph. Moreover #G = 2^n-#F where n is the number of variables of F. Thus, every known (structural) easyness and hardness result for the exact counting on CNF generalizes to DNF.

In other words, counting the number of model of beta-acyclic DNF is easy (from the paper you cite). And it is hard for alpha-acyclic DNF (just add a big clause containing every variables in a CNF). 

It is also hard for hypertrees: take a DNF F and construct F' by adding a fresh vertex x in every clause. #F = #F' and H(F') is a hypertree: take T rooted in x and attached to it every other variables as leaves. Observe that this reduction also shows that the problem #(k+1)-DNF restricted to hypertree is harder than #k-DNF so I doubt that you can compile them into OBDD. Moreover, taking k = 2 (already hard for counting), the transformation leads to a hypertree 3-DNF where every edge is a path in the tree so even this restriction is unlikely to yield easy cases.

DNF however differ from CNF as approximating the number of model is possible in the first case (see On the hardness of approximate reasoning Dan Roth - Artificial Intelligence, 1996 for the hardness of approximating and Monte-Carlo Approximation Algorithms for Enumeration Problems
Richard Karp, Michael Luby, Neal Madras).