Optocoupler min collector-emitter voltage
(1) Collector-emitter voltage: 80 V (min)


This is the highest voltage the transistor can withstand across its collector and emitter when in it's off-state.

The actual value varies from part to part. It will be at least 80 V for every device and may be higher for any particular part. (It probably will be higher, as manufacturers have to select a single rating value they can guarantee across millions of actual parts). Use 80 V as your design maximum, after allowing for tolerances in your circuit components and supply regulator/source.


  (2) Current transfer ratio: 50 % (min)


The ratio between the opto-coupler's LED current and the max. collector-emitter current. So an LED current of 5 mA would result in the photo-transistor conducting 10 mA max.

Again, the actual value varies from part to part but will be at least 50 % for every parts so use this as your design maximum.

...

You say your application circuit puts 5..20 V across the photo-transistor's collector-emitter when in its off-state and passes 5 mA through it in its on-state.

So the opto-isolator is suitable for the 5..20 V off-state voltage.

And the LED current will have to be at least 2.5 mA to conduct your 5 mA on-state current. However, I doubt that your load current is precisely 5.0000000 mA so you need to allow a suitable margin in LED current to cope with the max. actual load current. Then you need to ensure that this revised LED current will always be delivered to the LED by your circuit, allowing for supply circuit and supply voltage tolerances. Don't skip that stage.