Here goes...
To analyse the circuit, take C1 out of the circuit. R2 clamps the input of the op amp to ground. regardless of how much gain you have around the op amp (in this case 1000), the output of the op amp will be at ground.
the LDR has a voltage going to it from R1. If the LDR has a resistance of 10k, then the junction of R1 and LDR will be at Vcc/2.
If the LED shines through your finger any light coming out of the other side will be picked up by the LDR. The LDR will change its resistance according to the light it picks up (it goes down with increasing light level). As your blood pulses around your finger, the light level picked up by the LDR will change, causing the voltage at junction of LDR and R1 to pulse (sorry!) up and down.
You only want the pulses to be amplified (not the pulses + Vcc/2), so this voltage is capacitively coupled to the input of the op amp using C1. capacitors only pass ac and block dc (in this case they block the Vcc/2 voltage across the LDR).
The op amp then amplifies this (Pulsed) voltage. If you short it out, the Vcc/2 will be applied to the input of the op amp and the output will try to get to vcc/2 * 1000
Hope this helps