Electronic – Making a controllable H-bridge

You pretty much have the right idea about how the circuit works. When you pull Q10 down, you'll end up with Vcc across Vbe_12+Vbe_10 + resistors. Vbe's are generally around 0.7 volts, so you get Vcc-1.4V = Vresistors. Knowing the voltage over the resistors, you can calculate the current going through that path from VCC through the base of Q10. Once you know the current going through Q10_base with a known Hfe, you'll know how much current flows into the base of Q14. That current into Q14 will turn it on and its Hfe should allow plenty of current for the motor to flow through it from collector to emitter.

Raising the input high on Q10 causes the voltage drop on Vbe+Vbe+resistors to now be 0. Since there's no voltage drop, the PNP's are pinched off and no current flows through that path. Since no current flows through that path, no current is allowed to flow from the emitter to the collector of the PNP Q10. No current through that path prevents any current flowing into the base of Q14. Without any current flowing into the base of Q14, it prevents any current flowing from the motor to ground from that path.

Your third question falls somewhat in gray space. If you have matched transistors, then you can expect each complement to allow the same amount of current, you can expect the rise and fall times to be approximately equal and you can expect the ON-resistance of the transistor to be about the same. If you have a mismatch in this pair, you need to ensure that you're not compromising on any of those points. If one turns on faster than the other has time to shut off, you'll end up with longer times when both Q11 and Q14 are on at the same time. This causes a direct short often resulting in burned chips. If one transistor can't handle the amount of current that the other can supply, it will either prevent your motor from running optimally, or could cause your under-sized transistor to fail early. Basically, if you spec the non-complementary transistor correctly (ideally, overspec it a bit), then you shouldn't run into any issues.

You are not using a PNP transistor the correct way. With the base raised to 12V and the emitter as the output, the PNP B-E junction is being reverse bias. The BC557 B-E junction breaks down and current is supplied to the load through the break down. The TIP32C displays a different behavior probably because it breaks down at a higher B-E voltage.

With the description of your circuit, you want to use a NPN transistor.

^{simulate this circuit – Schematic created using CircuitLab}

Better yet, use a N-MOSFET. Since the gate of a MOSFET needs essentially zero DC current, the 12V signal does not have to supply a big BJT base current (order of 100mA) for the 2A output case.