Electronic – Why can’t I test the transistor part of an optocoupler using the multi-meter

Page 1 of the datasheet clearly says 'open collector output'. Without a pull-up, the voltage is undefined.

The purpose of 270 \$\Omega\$ on the test circuit is simply to provide a defined amount of current for the output logic to sink during the test. Whatever pull-up value you use is up to you, just don't exceed the sinking capability of the output stage.

When it comes to biasing an opto, the best practice is to choose a resistor that will allow the minimum amount of current needed to do the job over the expected current transfer ratio (CTR) distribution of the batch. Generally, the lower the current, the longer the expected life of the part.

No it's rather unsuitable. The 2us given on the datasheet is with a load of 100\$\Omega\$- which is a totally unrealistic load when used in a simple circuit.

If you allow a CTR of 10% (half the minimum when brand new) and put 10mA through the LED, you'll have 1mA at the phototransistor. A load of 10K would be reasonable, so delays will be in the tens of microseconds. See this datasheet.

It's not quite as bad as being proportional to the load resistance, but well in that direction, so the 100 ohms condition on the datasheet could, at best, lead to a bad conclusion. You could put a complex circuit around the phototransistor and get down to the datasheet numbers (which are not ouright lies, just misleading) and the base resistor you suggest will further reduce the numbers somewhat (at the expense of CTR, so a simple circuit does not benefit).

I suggest you use a logic-output optoisolator such as the 6N137 or similar. They use a photodiode internally with a sensitive and fast circuit, which is a much more suitable method for high speed. Optoisolators with phototransistor (or photodarlington) output generally should be avoided where speed is of much importance.