I'm looking at the datasheet of the ic and I kind of understand what's going on.
Apart from the typical sensing and signaling pins there are the HIDRV and LODRV pins which drive Q1 and Q2 respectively. (Typical Application, page 1)
Q1 controls the current flowing from a solar panel to the battery pack.
Wouldn't Q2 just drain the battery pack?
That's for the flyback currents. In simple converters, this would be a diode. However with a little more complexity inside the chip, they are able to use a FET instead, which reduces losses and improves efficiency.
The name of the topology is "synchronous converter". There's lots of info about them on the internet.
Here is an example of both topologies side by side:
Image from Maxim
I guess I understand what you're thinking. Q2 would actually drain your battery pack if only it were kept switched on for a time much longer than the actual switching frequency period.
But that's never the case.
This converter in the datasheet is called a buck converter, that is, it steps down your input voltage using the famous linear ratio \$V_{out} = Duty\,\cdot V_{in}\$. When Q1 is switched on, current through the inductor ramps up in a triangular fashion and it reaches a positive peak. At this moment, Q1 is switched off, and after the dead time (time between transistor switchings not to cause shot-circuits) Q2 switches on, and thus the current through the inductor now starts decreasing with a negative slope.
Now the ace in the hole comes into play, which is exactly the fact that the inductor current is indeed discharged, but it does NOT reach values below zero, that is, Q2 switches off and Q1 switches on at a moment in which the current through the inductor is decreasing yet positive, that means you'll not have your battery discharging through Q2.
Moreover, that is called a synchronous buck converter configuration because a diode would any losses to the circuit, and a MOSFET has a much lower \$R_{ds}on\$, decreasing power losses that would otherwise be added if a diode was the choice.
