Why does the phosphate ion have resonance structures?
A much better representation of phosphate is the one given below (You need to add three lone pairs to each oxygen atom which I was too lazy to do):




It does not expand phosphorous’ octet, which calculations have shown is unlikely.
It does not require phosphorous’ 3d-orbitals which are energetically too far removed to participate in hybridisation in a noteworthy manner.
It gives you four equivalent $\ce{P-O}$ bonds a priori without needing to resort to mesomery.
It gives all oxygen atoms the same charge a priori and also gives phosphorous a charge in better agreement with experiment/calculation.
The experimental bond lengths are better explained as single bonds rather than having a bond order of $\frac{5}{4}$.


This can all be summed up as ‘phosphorous obeys the octet rule much like every other main group atom (doublet rule for hydrogen/helium).’

But you probably want answers to your questions on the decidedly minor mesomeric representations your teacher drew. Well:



  Why should one oxygen be different than any of the others?


Those four mesomeric structures are entirely equal, just like in benzene. To get a ‘picture closer to the truth’ you would need to put them all on top of each other and look through them. You would see that every $\ce{P-O}$ bond has the same (minute) chance of being a double bond and thus no oxygen is different from the others.

  What diagramming rules imply that $\ce{PO4^3-}$ should be depicted as my teacher did?


Well: none, really. If you are going to stick with the double bond representation, the order of the four structures does not matter as long as they are all there. Also, you can go from every single one to every single other one in a single step by appropriate electron shifts. So, as MaxW commented, you really should add the two missing diagonal arrows to be correct.