It uses the large capacitor to "drop" the AC voltage down to a low value that can be rectified and fed (possibly via a resistor for some current limiting) to the battery. It's circuit might be something like this: -
The circuit above is possibly more complex than yours because it provides a semi stable 6.2V dc output (see the zener). I think your circuit will be similar in that there is a capacitor (2.2uF or maybe a slightly different value) that feeds the bridge then, I suspect, the battery will be where the 1000uF is.
The capacitor's impedance at 50/60 Hz will be a few kohms and at the battery's charging current will "lose" most of the AC voltage across it leaving maybe something like 5Vp-p going into the bridge.
The 470kohm resistor may be omitted on your circuit - it's used to discharge the 2.2uF cap when removed from AC - it could give quite a substantial tingle if someone handled the plug and the cap was still charged to a few hundred volts.