The circuit will output the set current if possible. If it can't reach that current, it will give as much as possible. If your load resistance is low enough, then it can deliver 10 amps, but that resistance will have to be pretty low. You will lose about 1 volt over Q2 (which is a darlington transistor, not two stacked on op of each other) and 0.7 volts over R1 (0.7V is a normal Vth) so if your supply is 9V, you have a maximum of 7.3V that can be supplied to the load. So if your load resistance is 1 ohm, it will never give more than 7.3 amps, even if you set R1 for a higher value.
If you just want to push a ton of current, this isn't necessary. I've heard some people say that delivering too much current to the igniter can melt the bridgewire so fast it won't heat the explosive before breaking. I'm not sure if this is true though. Other then that, such a circuit would only be good for limiting current consumption or powering current sensitive loads (i.e. glow plugs).