The first key, so they say, to understanding BJT behaviour is to understand that its driven by minority carrier behaviour. In an NPN device, that means that electrons in the p-type base region control the behaviour.
I think you captured that in your description, but most of the rest of what you wrote doesn't fit the usual way of describing the physics.
Since the base is very thin in relation to the collector and emitter, ... there are not many holes available to be recombined with emitter electrons. The emitter on the other hand is a heavily doped N+ material with many,many electrons in the conduction band.
This is the only part of what you wrote that makes sense. The forward bias on the b-e junction creates excess carriers in the base region. There are not enough holes to recombine with those electrons instantaneously, so the region of excess holes extends some distance from the beginning of the depletion region associated with the b-e junction. If it extends far enough, it will reach the opposite depletion region (for the c-b junction). Any electrons that get to that depletion region are quickly swept away by the electric field in the depletion region and that creates the collector current.
OK, so how is entropy involved?
A key point is that the spread of excess electrons away from the b-e junction is described by diffusion. And diffusion is, in some sense, a process that takes a low-entropy situation (a large number of particles segregated in one part of a volume) and turns it into a high-entropy situation (particles spread evenly across a volume).
So when you talk about "a high entropy of electrons", you actually have it backwards. Diffusion actually acts to increase entropy, not reduce it.
The idea that excess electrons are "effectively doping and shrinking the base/collector depletion region into N-type material" also doesn't make any sense. The excess carriers don't affect the extent of the c-b depletion region much. Electrons that reach the c-b depletion region are simply swept through by the electric field.
Your question is about the fundamental operation of a BJT. Surely there is much written about that out there.
Briefly, C-B-E is a sandwich of three semiconductors of opposite polarity doping. However, what makes a BJT more than just two diodes in series pointing in opposite directions is that the base region is so thin that the depletion region of each junction extends to the other junction. The collector is still within reach of the emitter, if it weren't for the base region in between with all its carriers depleted. A little bit of externally applied base current injects carriers into the base region, which now allows current to flow accross it between collector and emitter.
Due to a whole bunch of semiconductor physics you should look up elsewhere, a few carriers in the base go a long way. This is where the transistor gets its gain from. You inject a few carriers into the base (provide a small base current) and that lets a lot of carriers conduct (allows a larger current to flow) accross the otherwise depleted base region between collector and emitter.
Best Answer
The essence of transistor behavior in the active region is as follows.
Now:
The conductivity of the collector depends upon doping. Undoped silicon is a poor conductor.
The base is lightly doped to reduce the number of majority carriers in the base. The concentration of majority carriers is directly related to the doping concentration. The higher the concentration of majority carriers, the greater the chance of the minority carriers recombine in the base with majority carriers. This increases base current, and decreases collector current.
The collector is doped less than the emitter to make the transistor asymmetric. The more symmetrical a transistor, the poorer it performs in the forward active region (but the better it performs in the reverse active region).
Those are the reasons why the base and collector are both doped less strongly than the emitter. Why the base might be doped even less than the collector I do not currently know. Perhaps someone with more expertise in transistor design can answer.