long run a 12V mag or mech lock will be the load.
so why then does it say 1.35 is the typical max voltage?
Diodes can be very roughly modeled as a constant voltage load, meaning that you measure the same voltage across it no matter how much (or little) current is flowing through it. It is far from perfectly accurate, but for the ways LEDs are used it is usually good enough. 1.35V is the typical value of that voltage load for the optocoupler’s LED, when the forward current is 60 mA.
Because the model’s I-V characteristic is infinite (current can change infinitely with no change in voltage), an external component is necessary to restrict current. The crudest method is a resistor, though higher power applications (such as for illumination instead of indication) may use a more sophisticated current regulation circuit.
By consequence of Kirchoff’s Current Law (algebraic sum of all voltages in a loop is zero), you get the following equation with a resistor:
5V - IfR - 1.35V = 0
From there, simple algebra will allow you to choose an If value and calculate the resistor you need to get that. Or choose an R value and calculate the resulting If.