Working of op-amp based current source
The U1 is a current sense amplifier. It measures the voltage drop on R2 and it gives the actual current measurement to the U2. The U2 is a proportional controller with extremely high loop gain, since there is no feedback.
At the input you have a voltage divider 1:2 and this is a voltage setpoint, which is compared to R2 voltage (subtracted). The U2 shall output such voltage, so that V_R2 and V_in+ should match.
Now it's not very clear that you input 2V and you get 50uA. Let see: 2V is divided by 2, so the voltage on non-inverting input is 1V. The V_R2 would be 1V if the current is I=V/R=1/10k= 100uA.
EDIT:
My bad: The U1 measures the voltage drop on 10k resistor and has a gain of G=2, so correct, you get 50uA at V_in=2V.
To be able to get lower current you can amplify the signal from U1 by replacing R6 and R9 to a higher value.
EDIT2:
In order to get low current such as 1ua and little less, you would need to change the U1 with a JFET instrumetation amplifier, preferably with laser trimmed resistances. With actual schamtics, you have a current path from sensing resistor 100k to inverting input of U1. This is quite a lot of current passing with regard to your goals.