You have a photon that comes in and splits into $e^-e^+$. One of the electrons escapes, the other has a short segment, where it sheds another photon. The second electron now escapes and the photon is absorbed by the nucleus. The center electron segment and the second photon can be off the mass shell.
Quantum field theory does not offer a description of "how" its processes work, just like Newtonian mechanics doesn't offer an explanation of "how" forces impart acceleration or general relativity an explanation of "how" the spacetime metric obeys the Einstein equations.
The predictions of quantum field theory, and quantum electrodynamics (QED) in particular, are well-tested. Given two photons of sufficient energy to yield at least the rest mass of an electron-positron pair, one finds that QED predicts a non-zero amplitude for the process $\gamma\gamma \to e^+ e^-$ to happen. That is all the theory tells us. No "fluctuation", no "virtual particles", nothing. Just a cold, hard, quantitative prediction of how likely such an event is.
All other things - for instance the laughable description in the Wikipedia article you quote - are stories, in this case a human-readable interpretation of the Feynman diagrams used to compute the probability of the event, but should not be taken as the actual statement the quantitative theory makes.
There is no "how", what happens between the input and the output of a quantum field theoretic process is a black box called "time evolution" that has no direct, human-readable interpretation. If we resolve it perturbatively with Feynman diagrams, people like to tell stories of virtual particles, but no one forces us to do that - one may organize the series in another way, may be even forced to do so (e.g. at strong coupling), or one may not use a series at all to compute the probability. The only non-approximative answer to "how" the scattering processes happen in quantum field theory that QFT has to offer is to sit down and derive the LSZ formula for scattering amplitudes from scratch, as it is done in most QFT books. Which, as you may already see from the Wikipedia article, is not what passes as a good story in most circles.
But neither nature nor our models of it are required to yield good stories. Our models are required to yield accurate predictions, and that is what quantum field theory does.
Best Answer
A "Feynman" diagram does not have a time axis. This is, in fact, what distinguishes "Feynman" diagrams from the diagrams used in other accounting similar schemes. Every interaction point in a Feynman diagram is integrated over all possible space-time positions. Any set of vertices may be temporally ordered in every possible way (although obviously some permutations will contribute more to the total amplitude than others). Any internal line may represent a virtual particle (or antiparticle, for fermions) propagating in either direction between its endpoints. The reason for this convention is that it is (surprisingly) actually the easiest way to sum up all the amplitudes contributing to a given process.