Q1 is a common base amplifier. (non-inverting)
Often the Base is on ground. In this case the base is about 1 volt positive.
Usually there is a emitter resistor. If the collector resistor is 1ok (R1) and the emitter resistor is 1k the amp will have a gain of 10. If you put 10mV across the Emitter resistor then you will get 100mV across the Collector resistor.
If the emitter voltage is high then there will be little emitter current (maybe no current) so there will be little to no collector cur4rent so the voltage is high.
If the emitter voltage is low, in out case 0V then there is a large current in the B-E so large current in C-E and the C is at a low voltage.
Look at R4 & R2. The 3.3V point is not exactly 3.3 but reasonably stable. Most of the 3.3V is across R4, About 0.65 volts are in B-E of the Q2, and the rest of the voltage is on R2. About 0.25 volts on R2. So the Q2 emitter is at 0.25V. The base is one diode drop above that. Q1 Base is at the same voltage and its emitter is one diode drop below. So the two emitters are at the same voltage. If the temperature changes it should change in both diodes by the same amount.
You could use a diode in place of Q2. It will be close. Not the best. You could use just the B-E of Q2. That will be better. I used Q2 in what should be about the same way as Q1. If I can get aQ1 and Q2 to have almost the same amount of base current and the same collector current then the two Vbe will batch and the two emitter voltages will be the same. There will be no temperature problem based on Vbe.
How does Q2 work? Current flows through R4 into B of Q2. Q2 turns on and the collector turns on. The collector can pull below the base voltage. What happens is that the C sends most of the current to ground leaving a small amount of current for the base. Current is 2.3V/10k. Most of that current goes to the C-E and to ground via R2. The base current is just enough to get the C turned on for that current.
Q1 current is 2.3V/10k. Q2 current is 2.0/10k. About the same thing.
If the transistors have a current gain of 100, (at that current):
Q1 base current is 1/100 of its collector current. Q2 base current is 1/100 of its collector current. About the same thing.
The Vbe is dependent on current. Because Q1 = Q2 in many ways, the Vbe should be the same. Both Bases are 1 diode drop above the voltage on R2.
We have a high gain amplifier, (Q1). It thinks any voltage below 0.24V is low. Any voltage above 0.26 is high. There is a small linear region at about 0.25V.
The IC is looking to see if the voltage on FB is 1.25 volts. Because of Q1 this is translated to (is the voltage across the 0.7 ohms equal to 0.25 volts)? Current=350mA