Suppose we're interested in (minimally) coupling a heavy scalar particle to the electromagnetic field. In quantum field theory, we describe the particle as an excitation of a complex scalar field $\phi$. Including the simplest possible coupling leads to the theory of scalar QED, with Lagrangian $$\mathcal{L} = - \frac14 F_{\mu\nu} F^{\mu\nu} + |D_\mu \phi|^2 - m^2 |\phi|^2.$$ By expanding out the kinetic term, we see that there are vertices where a $\phi$ emits a photon, and where a $\phi$ emits two photons. The latter is required by gauge invariance.
On the other hand, if the particle is localized, we could also describe it as a worldline supplying a classical current. For a trajectory $y^\mu(t)$, the current is $$J^\mu(x) = e \int d \tau \, \frac{dy^\mu}{d\tau} \, \delta(x - y(\tau))$$ and the action is $$S = - \int dx \, \frac14 F_{\mu\nu} F^{\mu\nu} + \int dx \, J^\mu A_\mu + m \int d \tau.$$ Everything is gauge invariant, since $J^\mu$ is.
Both of these pictures are common, but I'm confused how to translate between them. On the quantum field side, we know that $\phi$ must be a complex scalar, and that it must have a two-photon coupling to ensure gauge invariance. But on the worldline side these facts aren't manifest at all! This doesn't make sense, because in both cases it should be possible to calculate, the probability of producing two hard photons when the particle is accelerated. The results should match, but since the worldline picture is "missing" the two-photon coupling, I don't see how it can.
More specifically, it's clear that the amplitude to produce one hard photon matches between the two formalisms -- what's puzzling to me is that it seems like this implies the amplitude to produce two hard photons in the worldline formalism should match the corresponding amplitude in the field theory formalism, without the two photon vertex.
What's going on?