Help explain how this ac switch circuit works
The source terminals of the two MOSFETs take on no higher than one body diode drop above the two drains when they are off.

When Q11 is off, R84 keeps the MOSFETs off.

When Q11 is on, the gates rises to near Vbat. Initially, when Vbat is greater than (one diode drop) + (Vce of Q11) + (Vgs-on of Q12) + (one body diode drop) + (the lower of the two drains), the two MOSFET would turn on. Since an MOSFET when on can conduct either way, they let current passes either way. The on-MOSFETs would try to pull the two sources and two drains to the same voltage. There is a possibility that the drain with the higher voltage can pull the sources higher to partially turn the MOSFETs back off. So for the switch to stay on consistently, Vbat has to be a certain level higher than the voltage levels which the MOSFETs switch.