Analog Comparator falling edge setup

The comparator(and C10) is essentially there

No.

Where should the µC know C from?

The meter you are trying to build does not measure ESR: it provides an indication of ESR, in conjunction with a known C.

Anyhow, can I improve my software solution or do I require two interrupts, without changes to the way the measurement is done?

Yes. This would be what I would do:

  1. measure the voltage on the capacitor, V0;
  2. charge up the capacitor for a known period of time, Tc;
  3. measure the voltage on the capacitor, V1; C = (V1-V0) / (I * Tc);
  4. discharge the capacitor for a known period of time, Td;
  5. measure the voltage, V2: V2 = V1 * exp (-Td / (ESR * C)), solve for ESR.

Done.

It uses the same hardware.