M-theory with its 3-form $H$ and the problem of having no Lagrangian

This is a standard question about the M-theoretic construction of the 6d (2,0) theory. This is done, in the simplest case, by an M2 brane hanging between two M5 branes. The theory on the M5 branes is the 6d (2,0) SCFT which has 4 fermions, 5 scalars and one self-dual 3-form $H=dB$.

My question is the following: why is one of the reasons of having no Lagrangian description for the 6d (2,0) theory the existence of the self-dual 3-form? Self-duality, in Lorentzian manifolds such as the world volume of the M5, appear in $d=2,6,10$. So what is the problem with it?

A related question is why is this theory not interpreted as a (higher?) gauge theory where $H$ is the (higher) curvature of the (higher) gauge field $B$? Usually people call this a gerbe.

The problem with a self-dual three-form field strength is that the obvious kinetic term vanishes: $$\mathcal{L}_\text{kin}=H\wedge *H =H\wedge H\,,$$ but since $H$ is a three-form, $H\wedge H=0$, so $\mathcal{L}_\text{kin}=0$. This can happen in $d=4n+2$, when you have a self-dual $(2n+1)$-form (in a chiral theory). Indeed, in type IIB string theory, there is a self-dual five-form field strength with the same problem.