The voltage across a capacitor is the integral of the current through it. If you feed a constant current to a capacitor, its voltage ramps up linearly, which is exactly what you want for a sawtooth waveform generator.
Yes, you're correct that this cannot continue forever; the complete waveform generator circuit will be discharging the capacitor periodically in order to prevent the constant-current circuit from saturating.
In the circuit you show, D1, D2 and R1 provide a voltage reference, in this case, approximately 1.2V, which is the total forward drop across the two diodes. This establishes the voltage across the B-E junction of Q1 and R2. Since the voltage across the B-E junction is also a diode drop, or 0.6V, this means that the remaining voltage, 0.6V appears across R2.
From Ohm's law, we now know the current through R2, 0.6V/2200Ω = 273µA. This is the current that is charging C1.
The voltage across the capacitor is a function of time: V = I×t/C. Let's rewrite this as V/t = I/C, which means that the rate of change of the voltage is the current divided by the capacitance. In this case, 273µA/0.1µF = 2730 V/s, or equivalently, 2.73 V/ms.
The output capacitance of the IRFH6200 is 2.89 nF with 10 volts from drain to source. With zero volts from drain to source this capacitance will be in the realm of 10 nF so, when you apply the pulse to VL, irrespective of how long or short it takes for the control system to recover, you are (in effect) applying that same pulse directly to RL via a capacitor that starts at about 10 nF and diminishes to maybe 2 nF when the peak of the pulse is reached.
What happens is, that there is a large unrestricted peak current
flowing through RL for about 20µs
That initial current flow is due to the capacitor and the fact that the MOSFET, prior to the injection of the voltage pulse is hard-on due to the op-amps. How long it takes for the control loop to stabilize is down to choosing a faster op-amps (in part) but also it's down to being able to remove charge from the MOSFET gate (power drivers are usually rated in amps to accomplish this).
The gate source capacitance is about 10 nF and a measly 20 mA from an op-amp is going to reduce the gate voltage at a rate of 20 mA/10 nF = 2 volts per microsecond - it might take 5 to 8 us just to turn the MOSFET off.
Best Answer
I can speak from some small experiences on a part of this. I'll leave the "highly constant and highly stable" current source design to those who know this better than I. But I was involved in creating LED-based light sources intended as "lamp references" for optical work. Your description suggests the same to me.
(1) Even if you have a perfect current source, LED output varies with temperature. So they must be operated at a stable temperature. We chose to heat them to about \$75^\circ\text{C}\$ using a closed loop control. That solved the temperature drift problem.
(2) LEDs themselves aren't consistent. Even those cut from the same wafer aren't consistent. We had to pre-stress thousands of them, holding them at a stable temperature while running them with a current source for days. We monitored the optical output and data logged the results. A few, a SMALL FEW, would actually gradually drift and then settle down to a stable spot (peak wavelength and intensity.) We chose those for the standards. We threw away most of them, by far. At the time, we kept less than 1% of those tested because more than 99% of them simply weren't stable even over periods of hours.
I'm just saying. I believe you are focused on the wrong issue. Buy yourself a really nice 100ppm "stability" current source from a supplier that provides the range and resolution and accuracy you want and guarantees all of it (it won't be cheap.) And then use that with your LEDs and play around with temperature and various LEDs just to sample the modern situation with LEDs before going off and worrying about the current source end of it. You need to know what parameters need to be placed under your control in order to meet your requirements. Dog first, tail later. Just one opinion.