The Base-Emitter junction is basically a diode, a PN-junction. It's resistance changes with applied voltage. So, if you were to raise the voltage at a constant speed, the current wouldn't rise at a constant speed, it would curve up fairly quickly. Because diodes will have lower resistance at higher voltages, the voltage drop across them will actually stay pretty constant, think of resistor voltage-dividers. So, a rough estimate of current through a diode (or Base-Emitter) is to take the input voltage, subtract the "forward-bias" voltage (the voltage that the diode will PRETTY MUCH drop, regardless of current), and divide by the resistance of the rest of the circuit that's in series.

Now, the Collector-Emitter junction will do it's best to adjust its resistance in such a way that the current passing through it is equal to the current through the Base-Emitter junction multiplied by Beta (which varies considerably in most BJT transistors, can be anywhere from 10-500, depending on different factors, usually its closest to 100). So, if you calculate the current through the Base-Emitter, you know how much current will TRY and pass through the Collector-Emitter. If the Collector-Emitter junction can't pass as much current as it wants (like if you have a resistor leading to it, and that resistor can't possible pass that much current given the source voltage) then the transistor will do the best it can, and reduce its resistance to effectively zero. This is called saturation.

Try drawing some circuits and fiddling with the mathematics a bit, it'll make more sense then.

There are other operating regions, but that's the one I use and understand the most.

Can anyone check me on everything I said? I'm still hesitant to act like an expert here on the forums

 

You are not going to think that this is a useful post, but here goes, obtain one of the many books on junction transistors and read it. Then come back with questions. This was you will get partial information and may never come to a true understanding.