How do I find out power requirements from AC / DC specs?
Power = V x I x PF

Where PF = power factor. For "normal" loads (e.g. resistive) PF = 1, so Ohm's law in its most familiar form applies. That covers all the typical educational introductory laboratory test cases so we start there to avoid confusing people needlessly.

For your first device...

You can tell 2 pieces of information (approximately) for your first device:

1. Efficiency

Efficiency is broadly defined as out/in. Here, Pout/Pin. 
Pin,min = 100*2*1; Pout,max = 24*6.25*1; Ergo, the best possible efficiency is 75% because we assumed the least amount in for the most amount out at the best power factor (1).

2. Maximum Safe (supposedly) Output Power

Use your original concept of an assumed PF=1. Volts * Amps * PF = Watts. Interestingly when you are assuming PF=1, rather than knowing/claiming PF=1, the unit used is called Volt-Amps (VA) to illustrate this. If you google for consumer UPS (battery backup), you will observe that they are almost universally rated this way.

So a valid conclusion is that this unit can output up to 24*6.25=150VA.

For your second device...

If the specs can be believed (always test if it's mission critical). They are claiming that the device is reasonably efficient and the output voltage is either known (indicated somewhere else) or adjustable (range should be published somewhere). 

Therefore, the output current is not as important as the output power since there is a relationship between output voltage and current that has a (roughly) fixed product (power). They have specified the input current, voltage, and frequency, but the output power. 

Incidentally, the numbers look a bit off to me since input power is >150W and output power is only 60W... Could you post a photo of the label?