on average they form dipoles.
Not quite. On average they form the nice (electrically neutral) electron orbitals we know and love. However, when they happen to be in a dipole state, there is an attraction between the two atoms, and that is evidently more than enough to overcome the randomly occurring opposing dipole states. That's just mathematics.
why is the potential energy of a dipole lower than a non-dipole
It isn't. But quantum mechanics allows electrons to visit higher energy states, with a probability dependent on the energy difference.
What determines which electrons are going to move to the middle of the atoms to form a dipole, and which ones move to the outside of an atom
Nothing. The "movement" of electrons is just a random statistical fluctuation in the location of the electrons' centers of charge. Quantum-mechanical events like this are inherently nondeterministic. And strictly speaking, you can't tell electrons apart anyway. There is no "which electron," at one point in time you have one electron here and another there, the next they may have switched places. Or not, you can never know.
(I assume that a dipole is formed like this $e−..ion...........e−..ion$ for the centre of positive charge shifted rightwards of the centre of negative charge)
Don't think of those as being electrons, just the statistical average of the negative charge distribution.
given that the jitters are instantaneous, and photons travel only at the speed of light, how are dipoles formed quickly enough before the next jitter
First, They're not quite instantaneous (that would have them basically everywhere possible at once), and not quite jitters.
In summary, don't think of electrons as particles with a distinct position moving around, just think of them as waves, with a peak that occasionally forms, and that peak causes the electric arrangement of two nearby atoms to, on average, be mildly attractive.