Why feedback is activating at more than 4V?
Your curve is actually composed of 3 parts. 1 is from 0 to ~3.7 volts, with essentially zero amps. 2 is from about 3.7 to about 4.3 volts, and in this region current is essentially linear with respect to voltage. 3 is above 4.3 volts with a gradually limiting increase in current.

1) occurs with the NPN shut off due to 0 Vbe, and lasts until the supply voltage, through Rin, reaches the MOSFET gate threshold - in this case about 3.5 to 3.6 volts.

2) Further increase in supply voltage causes a rapid increase in diode current, which also produces a rapid increase in base voltage in the NPN, but not enough to turn it on.

3) When the supply voltage hits about 4.3 volts, diode current reaches nearly 200 mA. This produces about .5 volts on the NPN base , which is enough to start producing collector current  and reducing the rate of increase in the gate voltage. It starts at this low base voltage (less than the usually mentioned 0.6 to 0.7 volts) because Rin is so high that it takes very little base current to produce an effective collector current. As the supply voltage continues to increase, the transistor turns on more and more strongly, limiting the increase in gate voltage and therefor in base voltage.

To check this, add a circuit consisting of a MOSFET with gate voltage tied to your M2 gate, source to ground, and drain to DC supply by way of a 1k resistor. Monitor the drain voltage of the new MOSFET. You will see that the drain voltage drops to zero just before the diode current starts to increase.

And for what it's worth, the reason you don't see much change with Rin is that the NPN transistors you're using to simulate this have very low leakage current. If you try an older transistor such as a 2N2222, you'll see a much higher turn-on voltage. This is because leakage in the NPN acts as a voltage divider and reduces the M2 gate voltage. To check this, monitor the gate voltage of the MOSFET, and you will see that for your initial NPN the gate voltage closely follows the supply voltage until the NPN starts conducting. With an older, leakier NPN you'll see the gate voltage significantly reduced.

So the simple answer to your post question is: Because 4 volts is the sum of the MOSFET gate threshold and the NPN base conduction point.