You sure managed to both oversimplify and over complicate it in one go, without getting it completely right. When you copy something out of a school book or internet text, you really should add your own "filter"
Regarding BJTs, the following summarises operation for an NPN BJT:
If the base voltage is greater than both the collector and emitter voltages and greater than the emitter voltage by 0.7V or more, it is in the saturation region. In this region the voltage across the collector/emitter is about 0.2V and the device acts as a closed switch.
First, the collector is at what voltage it is, as a consequence of the base voltage (or rather currents, but luckily, Ohms law is valid here as well, so lets just stick to voltages here).
Your first statement means that if emitter=0V, base=0.7V, the collector can anywhere up to just under 0.7V, but your next statement says that the collector will then be at 0.2V - seems you're in disagreement with yourself here?
The 0.2V c-e drop is pure school book though, in the real world, a saturated transistor can be from 0.045V (45mV) to close to 1V (and I'm not referring to Darlingtons here).
The b-e voltage drop can vary some as well.
If the above is true except the base voltage is less than the collector, it is in the active region. In the region, the collector current is proportional to the base current and the constant of proportionality is betabwhich is specified for thr device.
Not entirely so - see below
If the base voltage is less than the emitter and less than the collector, or less than 0.7V greater than the emitter, the device is in the cutoff region and the collector current equals zero.
Again, the collector is at what voltage it is, as a consequence of the base voltage and as long as the base is less than the required voltage drop to get the transistor to conduct, it will be less than the collector, so it's superfluous info and just confusing the text to mention the collector and the "less than the emitter".
And it could be expressed short, with precision and no ambiguity by:
BjT NPN, emitter at 0V:
If the b-e junction isn't forward biased (1), the transistor is in cut-off (not conducting).
If the b-e junction is forward biased, while the b-c junction isn't, the transistor is in it's linear region.
If both the b-e junction and the b-c junction are forward biased (and Ic/Ib < Beta), the transistor is in saturation.
1) The forward bias voltage needed is usually
around 0.6V to 0.7V for small signal work on silicium.