The process

Twice now you have drawn one diagram and squared it. Lesson 2.1 annihilated an electron and a positron into a photon that made a muon pair; lesson 2.2 sent an electron and a muon past each other, trading a single photon. Now point the electron and the positron at each other: $e^+e^- \to e^+e^-$. The same two particles come out. This is Bhabha scattering.

Something new happens here. There are two ways the reaction can run. The $e^+e^-$ can annihilate into a single photon that immediately re-creates an $e^+e^-$ pair — the 2.1 diagram, wearing a different hat. Or they can stay themselves and simply exchange a photon — the 2.2 diagram, with a positron where the muon used to be. Both begin with $e^+e^-$ and end with $e^+e^-$, so both are the same reaction.

That is the whole lesson in one line: when more than one diagram ends in the same place, you may not square them one at a time. You add the amplitudes first and square the sum — and the extra piece you get is interference. Before any of that, find the diagrams. Draw every way this reaction can happen.