Confusion with transistor modes

First, lets look at why there's only very small current in a reverse-biased pn-junction diode. The junction doesn't block all current when reverse biased. The electric field in the junction opposes the majority carrier current whether forward-biased or reverse-biased, but quickly sweeps any available minority carriers (electrons in the p-region, holes in the n-region) across it. In forward bias, minority carriers are being continuously injected from the contacts, so there is a sustained current. In reverse bias conditions, there's very few minority carriers available, so the junction carries all the available carriers away in a very short time, and there's no more carriers available to sustain a current.

So what happens in a forward-active BJT is that the forward biased base-emitter junction creates a large number of minority carriers in the base region. The (reverse-biased) collector-base junction then has no problem "finding" carriers to create a current, and so you can have a large collector current.

It is not correct that the depletion regions of the two junctions overlap. If that happens, you have a condition called "punch-through" where there is no gain in the device.

I found a slide-set that gives a very quick explanation of BJT operation here. In particular note that current in the depletion regions is mainly caused by "drift" (carriers being pushed around by electric fields), but in the bulk regions it is mainly caused by diffusion --- that is simply carriers randomly moving around, so that the net movement is from areas of high concentration to areas of low concentration. Finally, remember that the important currents are the minority-carrier currents.

Edit

My explanation of forward biased operation was not right. Let's try again: Whether the junction is forward or reverse biased, the electric field in the depletion region (the area right around the junction) opposes "majority" carriers crossing the junction and encourages minority carriers. In forward bias, the size of this barrier is reduced to the point where some fraction of the majority carriers have enough thermal energy to overcome the barrier. But anyway, the operation in reverse bias is more important to answering your question.

1) The essence of an NPN transistor is that it is not just 2 PN diodes in anti series (like NP - PN ) but an NPN structure where the P-layer is quite thin.

So what happens is that the electrons coming from the emitter (at some speed) while traveling through that very thin base, have a large chance of going too far and ending up in the collector. This happens despite the base-collector junction being reverse biased.

The thing to remember: thin base !

Also, the doping in the base is also such that the base cannot "catch" (recombine) all the electrons so many electrons make it "safely" through the base without recombining (with the holes in the base) and end up in the collector. I hope you realize that having a thin base region helps as well.

2) You mean, leave the base floating.

I only discus the majority carrier currents here. There is also a much smaller minority carrier current, see this question and the answer Since the effect of this is much smaller, I ignore it here.

When the base is floating, the Emitter-base junction cannot come into forward mode. There would then be no electric field to accelerate the electrons from the emitter so they cannot make it into the collector.

So the NPN will not conduct (between collector and emitter) in this case as the electrons from the emitter cannot make it to the collector as their energy is not sufficient. They need the base to be at a higher potential (the base-emitter junction must be in forward mode) in order to accelerate/gain energy to make it into the collector.