Physical operations of discrete - time circuits
While phase 1 is active, C1 charges to the level of the input, and retains the last voltage it sensed when phase 1 ended.  When phase 2 is active, the negative of that voltage is applied to the op amp input, forcing the output to change until the input is at ground.  This discharges C1 and charges C2.  The resulting voltage will stabilize at Vin times C1/C2.  When phase 2 is over, C2 maintains its charge, so the next phase 2 will add the input cumulatively, hence the integration.

Another way to look at it (depending on which is easiest for analysis) is that during phase 1 C1 charges up, and during phase 2 the charge is transferred to C2, in addition to any charge already on C2.