Energy and Momentum conservation in Bhabha scattering I came across this question here which was asking my exact initial question (I even came from Griffiths problem 2.4 too!). But the given answer gave me a lot more questions which I think constitute too much content to fit in a comment, so I hope this warrants its own question here.
To summarize, we have Bhabha scattering (with time flowing left to right):

As the accepted answer puts it, we have energy and momentum conservation at each vertex, as well as overall. Thus for each vertex we have $E_{e,\text{in}} + E_\gamma = E_{e,\text{out}}$, and since $E_{e,\text{in}} = E_{e,\text{out}} = m_e$ (in natural units), we have $E_\gamma = 0$.
My primary question is: how do you determine the direction of the photon (alternatively, which vertex "happens first")? The answer on the linked question seems to imply they happen simultaneously; but in that case, how do you know which side of the energy conservation equation the $E_\gamma$ goes on? In this case, it of course doesn't matter either way since $E_\gamma$ is 0, but I could imagine interactions where this does matter.
Also, were it not for total energy conservation (say we had the identical diagram for the invalid "scattering" $\mu^-\mu^+\to e^-e^+$), I assume the photon in fact could take on energy. But if one vertex happens first, could the the photon not take on a variable amount of the excess energy $m_\mu - m_e$, with momentum conservation then being ensured by the second vertex? I have only ever seen variable energy to be integrated over in diagrams with loops, so it feels like this intuition is wrong.
Hopefully this isn't too many small little questions; help is appreciated.
 A: Feynman diagrams are a diagrammatic representation of the terms needed to calculate interactions of elementary particles. What is really measurable in experiments are the four vectors inputed, and the four vectors going out. The drawing and labeling the internal line as a photon is only done for a mnemonic representation of the quantum numbers exchanged during the interaction. The four vector  of the line is "virtual". Example

This means that the four vector of the "virtual  particle" does not have its mass, it is off mass shell and within the complexities of the diagram can have any value that is allowed by the conservation laws of energy and momentum. The four vectors of real particles have as length the mass of the particle.

how do you determine the direction of the photon (alternatively, which vertex "happens first")?

In Feynman diagrams the axis of the variables are defined. In the Bhabha case the horizontal axis implies the same time for the two vertices. Considering that experimentally we do not see first one and then the other real particle with a time delay, the time axis in the diagram is just a coordinate confirming this.

But if one vertex happens first, could the the photon not take on a variable amount of the excess energy mμ−me, with momentum conservation then being ensured by the second vertex?

As I state above, all experiments show simultaneous time for the exit particles. Energy can be carried by the virtual particle, example:
In Compton scattering  the virtual particle is a virtual electron, ( see diagram here )its four vector will be off mass shell, but there is energy transferred in the interaction.
