The answer is that it varies with the needs for the design. You can use resistors to bias an AB class amplifier (just look for Class AB on Google images to see a huge variation in designs), though typically you will see some other schema for setting the bias current through the output pair, or a constant current drive for the bases.
A lot depends on whether you are setting very low class A capability or not. When using two diodes to bias the output pair you are operating very close to Class B only.
In the schematic below it shows both low and high A bias:
What you describe with two diodes (which have Vf close to the Vbe of the output pair) are operating at very low Class A current. Typically you might see this in a 5-10 W or so amplifier. Operation in Class A (linear) may only be 200-300 mW. These amplifiers turn up in battery powered products a lot since they have low bias current
In the second circuit there are 4 diodes, and you'll see this commonly in high power (50-100 W) mains driven amplifiers. Here the linear operation may cover 5 W or more. This is done so that when you plug in a headset you typically only use the output as a Class A stage with very low distortion.
The two diode bias obviously tracks better thermally than the four diode amplifier, but the larger amplifier is more capable and the output stage has larger heatsinks.
In terms of understanding this type of amplifier you can do no better than read the Linsley amplifier design, done before the days of FETs.
Equally great reading is Blomleys 'New Approach' design.
These guys were at the forefront of amplifier design in the 70's, but FET based designs changed it all.
That looks like a good design, but I see two issues. One is that the typical idle current for an audio amplifier is 50 mA from emitter to emitter. Once they warm up you adjust the bias resistor again to get it back to a stable 50 mA. That is enough idle current to have low distortion, less than .01%. By the way that would be 50 mA per each pair of output transistors, so you can imagine the idle current of a 1,500 watt amplifier with ten output pairs.
The output emitter resistors vary from .15 ohm to .47 ohm, with .33 being a common value. 0.33 ohms creates a .65 volt drop at 2 amps of current, enough to trip NPN and PNP current limiters. They may cause clipping at high volume, but they will protect the output transistors from excessive currents.
Look at any op-amp design (LM324) and you will see those current limiters in place and insert yours accordingly. 0.33 ohms limits it to 2 amps or about 70 watts RMS. 0.15 ohms doubles the current to 4 amps. Put two .15 ohms in parallel if you really think this can handle ten amps. Do not exceed 50% of the transistors max amp rating.
The second problem I see is that there is a possibility of over-voltage at the IC (+) input pin. It is powered by 12 volts, yet if that 47 K potentiometer is set to high it will put more than 12 volts into the IC, which would destroy it. You CANNOT have an input with a voltage higher than the power pin, or more negative than ground.
Anything is possible, but you need to heat sink those transistors with a beefy heat sink and maybe a cooling fan if your going to push things. The .65 volt drop is for add-on generic NPN (2N3904) and PNP (2N3906) transistors, which short the base of your output transistors to the common output line. Also 10 nF may be a bit high for a feed back loop. They are usually 22 pF or so.
Try and see if the 47 K pot can be tied to the 12 volt rail, thus avoiding a chance of blowing the IC.
Best Answer
Well.. one is current controlled and the other is voltage controlled.
Another point to make is that it's very easy to open a BJT half-ish, with a MOSFET you got an equation that looks a little bit like this: \$I_{DS}=K×(V_{GS}-V_{TH})^2\$. There's a square in there. For the BJT it's just \$I_{CE}=I_{BE}×\beta\$ (in saturated region). For the MOSFET it means that when the sound goes up, on the output you get the square of that... which is not really what you want, you want to linearly amplify the sound, like the BJT does.
You want this \$Y=K×X\$, not \$Y=K×X^2\$
If you however use some feedback then you can make it linear... like using an op-amp. Most modern op-amps, especially those of modest specifications, will be MOSFET-based, if you just make a feedback with resistors then the amplification will be linear, which is what you want, and then it will work nearly identical to the class B amplifier.
This is the schematic of some MOSFET based op-amp. It's too much hassle needed to linearize a class B amplifier.
But as you can see, the \$M_{PO}\$ and \$M_{no}\$ forms a class B amplifier stage, there's just so much fuzz in the foreground to linearize it. For an op-amp where everything is in a small package, no problem. All the transistors receive roughly the same \$K\$ since it's all in one batch. If you however need to use all the individual components in packages then there will be some mismatch with the \$K\$'s (in the equation above) because the components were all made in different batches. With different \$K\$ in the above equations, everything will be... mismatched and you'll end up with a dysfunctional op-amp which will make your class B amplifier dysfunctional.