How is this ground fault circuit working?
The circuit is just a peak detector, that has a gain of 100 with a little bandwidth limiting Low Pass Filter, followed by a comparator that compares the output of the peak detector with 0.8V.

C14 is not for a delay, its there to set the peak hold time.  It's charged through the diode, D3, but has no discharge path so holds the highest voltage it sees. Normally there would be a parallel large value resistor to make this hold time more predictable, currently it's set by the comparator opamps input resistance and it's own leakage.

As the current in the CT increases (because of a fault) the voltage across R17 increases, this is clipped by D1. The peak detector amplifies negative signals better than positive ones (because it's inverting and the output can't go below zero), so it smooths out the bumps in the negative AC, as well as (for a short time) capturing the peak voltage of the negative cycle.  If this peak exceeds 0.8V the output goes high.  

The circuit is missing a latch to capture the fault signal, so as it is it would oscillate.  When the load is remove by the detector circuit, the current would drop to zero and after a short time the fault output would be de-assert, turning back on the load.