First let's clear up some of your misunderstandings. This will take some time.
Doping does not change the net charge. For every phosphorus atom introduced, a free electron is introduced. But also, an additional (compared with silicon) proton is introduced. After doping, the silicon is still electrically neutral.
Doping does change charge carrier density. In the N-doped region, four phosphorus valence electrons bond with four nearby silicon atoms, and the phosphorus locks neatly into the crystal lattice. The remaining electron is (must be) elevated to a higher-energy state, where it is essentially unbound to the atom that donated it, and therefore it's "free" to move around. Like I said, its charge is cancelled out by the phosphorus' additional proton, but the electron is free while the proton is immobile (stuck inside the nucleus).
In a similar fashion, when doping with boron, the boron locks neatly into the crystal lattice despite having one less electron than it ought to. There is then a vacancy (a hole) in the valence bonds it forms with its neighbors. This hole can be filled by a neighboring electron, but that then opens a vacancy where that electron came from. In this way, doping with boron introduces charge carriers that are most conveniently thought of as holes, which have a positive charge. A hole isn't a "real" subatomic particle, but it's convenient to work with mentally. (It's much easier to think of a single hole moving about, rather than the large population of electrons that move one space over trying to fill it!)
When a P-doped region touches an N-doped region, a "built-in voltage" forms: the N-region becomes positive, and the P-region becomes negative. I'll say it again: The N-region is POSITIVE, and the P-region is NEGATIVE. This seems backwards at first, but let me explain.
Remember that both N- and P-regions started neutral. When they touch, the electrons and holes shouldn't go anywhere, because everything's neutral, right? Wrong. On the N-side electrons have high density while on the P-side they're very scarce. Because of the imbalance, a process called diffusion carries electrons to the P-side.
Diffusion is the process that carries a fart across a classroom. Thermal (heat) energy in the air jiggles and jostles the fart molecules around in random directions. Each molecule takes a path called a drunkard's walk. On average, the fart molecules tend to go from regions of high density toward regions of low density. This isn't because they repel each other or anything; it's just a product of randomness. With enough time, thermal energy diffuses the fart until it is uniformly spread throughout the classroom.
Electrons would become uniformly spread too, because of random collisions with the crystal lattice, if it weren't for their charge. When electrons diffuse from the N-region to the P-region, they carry with them negative charge (and leave behind positively charged nuclei, or positive ions). Thus, the P-region becomes more negative, and the N-region more positive. Eventually a voltage forms which arrests further diffusion, and this is called the built-in voltage. It's usually about 0.8V. All P-N junctions have a built-in voltage inside, although you can't measure it with a multimeter because as soon as you touch the silicon a contact potential forms which cancels it out.
I should note here that the same thing happens with holes: holes are in abundance on the P-side, so they diffuse to the N-side, making it more positive while leaving behind negative ions on the P-side.
It's the built-in voltage that opposes current flow. Reducing that voltage by applying a forward bias allows electrons to diffuse from the N- to the P-region again, and likewise it allows holes to diffuse from the P- to the N-region. Remember, the built-in voltage makes the P-region more negative. Making it more positive again reduces the barrier for electrons to go there. Which some of them will do, simply because thermal energy launches them there.
So in an NPN transistor, in order to get current to flow, the voltage of the P-region (the base) should be made more positive than the emitter.
Another note: the current is not proportional to the voltage, but is instead exponentially related to the voltage.
Once in the base, electrons randomly wander about until they get collected. After forward-biasing the base-emitter junction, electrons diffuse into the base region, and then they wander about. Remember that the collector is at high voltage, so any electrons that manage to get there will [almost instantly] fall into it. (Electrons love going toward higher voltage.) Therefore a concentration gradient is set up in the base region: zero concentration at the base-collector junction, and non-zero concentration at the base-emitter junction. This gradient causes a flow of electrons from emitter to collector. Again, it's entirely a random, thermal process.
Now... Let's get to your questions. I'm sure you'll be brimming with more, but everybody starts somewhere.
Since regardless of polarity there is an "N" slice on either side of the "P" slice, why do BJTs have polarity? What is the difference between the collector and the emitter?
The emitter is intentionally very heavily doped. This means that when you forward-bias the base-emitter junction, more electrons will diffuse from N to P than holes from P to N. This improves the transistor's \$\beta\$. However, it also means the base-emitter junction has a very low breakdown voltage (typically around 7V). Because the collector is not intended to inject electrons into the base, it can be made lighter-doped to permit a higher breakdown voltage (e.g., 100V).
On a datasheet, where does one find the "ratio" of charge of "P" to the total current allowed to move through the transistor?
The transit time dictates how much charge must be stored in the base to obtain a given current. Namely, if it takes on average \$\tau_B\$ seconds for an electron to travel from emitter to collector, then the charge stored in the base to obtain a current of \$I_C\$ must be \$Q=\tau_BI_C\$.
The transition frequency \$f_T\$ is often listed in datasheets, and you can find transit time by \$\tau_B=\frac{2\pi}{f_T}\$.
A bit of extra charge must also be invested to charge the depletion capacitance.
When we apply negative charge to "P" and current is allowed through the transistor, is the current being applied to "P" (minus the charge required to overcome "P"s positive charge) along for the ride? I.E is the current applied to be added to the current on the emitter?
Whatever current that flows into the base does indeed flow through the emitter (and not the collector). Thankfully this is usually a small current, somewhere around 1% of the current flowing in the collector.
When used in a circuit, an NPN transistor can only be applied to the current path which is electrically common with the negative terminal of your power source, correct?
I don't know what you mean by this.
