I'm sure there are some opamp tutorials around, but I couldn't find a good one atm; hopefully someone can locate one for you.
Basically, for an opamp with negative feedback (resistor going from output to -ve input), assume the following rules (these are simplifications that allow you to work out what's going on):
1. the -ve & +ve input will be at the same voltage (the opamp will change its output voltage to cause this to happen)
2. the -ve & +ve input draw no current (they have very high impedance)
So the top opamp U1A has 0.47V applied to it's +ve input. It also has negative feedback through R1 & Q1, so the above 2 rules apply. So the -ve input is held at 0.47V too; this voltage is also across R2, which we can calculate to be passing 0.47V/4.7ohms = 100mA.
So, this opamp circuit will ensure 100mA is flowing through R2. A similar current (ignoring Q1 base current) will be flowing through the Li Cell.
The bottom opamp has positive feedback, which causes the output to be either high (~4V) or low (~0V); you can't use the above 2 rules for this one and it works like a comparator (i.e. when -ve input is less than +ve input, output is high; else output is low). When the output is low, the voltage divider R4 & R5 set a level of 1.58V (assuming 1.7V at V SET) on the +ve input of U1B. As the Li Cell charges, the voltage on the -ve input decreases until it passes 1.58V which causes output to go high. This turns on Q2, which turns off the constant current source. As the output is high, the voltage on the +ve input is increased to 1.8V (due to resistor divider).
As the Li Cell voltage discharges (for whatever reason), the voltage on the U1B -ve input increases. When it passes 1.8V, the output flips to become low again. Q2 is turned off, and the constant current is reapplied.