Output stage amplifier bias

Bipolar junction transistors (BJT) have 3 primary modes of operation:

1. Off: no significant current flows from the collector to the emitter
2. Saturated: the transistor behaves like a diode (which is why there's a fixed voltage drop across the collector/emitter pair under saturation)
3. Linear: this is where the transistor behaves like a current amplifier. The current flowing from the collector to the emitter is proportional to the current flowing from the base to the emitter. Note that in this region the voltage drop across the collector/emitter pair is NOT roughly constant, but linear.

In cases 2 and 3 there is current flowing from the collector to the emitter.

You can saturate the BJT by attaching a smaller load (i.e. higher resistance) than the linear limit allows. Think of a pipe with a valve. If there's something clogging the pipe upstream, at a certain point no matter how much you open the valve you're not going to get any more fluid flowing through the valve.

Yes, the electrons do literally move to generate the current, though this isn't directly used for most electrical engineering analysis.

For most EE analysis we don't consider the motions of charged particles directly, instead relying completely on the abstract concept of current. This is an approximation of reality, but it turns out to be remarkably good in most cases.

Ironically, in my day-to-day work with plasma physics we do consider the motion of charged particles instead of relying on abstract currents because the analysis does differ depending on what species are moving and where/how fast they are moving.

The 0.4 V across the Thevenin resistor (R1//R2) is a somewhat arbitrary design choice based on the emitter voltage. It is the voltage drop caused by the Ib of the transistor because it loads the voltage divider made by R1 and R2.

Choosing more than 0.4 V would result in higher values for R1 and R2 resulting in a larger voltage drop caused by Ib resulting in ending up with a different emitter voltage than the 4 V we wanted.

And likewise a lower value would result in less voltage drop.

I think they just used 10% of the emitter voltage (4V) as the value. That makes sense as the 0.6 V drop from \$V_{be}\$ is relatively constant. And if the emitter voltage was lower, 1 V for example, you would want the voltage drop due to \$I_b\$ to be smaller as well to prevent ending up too far away from the initial calculations.