Why is stepping up voltage during transmission considers I^2*R but not V^2/R?
For these transmission lines, not the voltage will result in a power loss, but the voltage drop on these lines.

I thinks its easiest explained on an example:
Lets say your transmission line has a resistance of R=100Ohm and you want to transfer P=1kW.

With "P=U*I" you get:

@1000V, you need to transfer 1A

@100kV you need  to transfer 0.01A

By "dU=R*I" the voltage drop across your transmission line will be:

100V @1000V

1V @100kV

As the voltage drop in your transmission line is the power lost you can now calculate the Power loss by "Ploss = dU * I" which results in:

100W @1000V which is 10% of your original 1kW

10mW @100kV which is 0.001% of your original 1kW

Thus: The higher the voltage, the lower the power loss in your transmission line (which of course also has it's upper limits due to for example isolation of these lines and isolation of the transformers, but thats another topic).