How does a capactior reduce voltage ripples
The AC source acts as an independent voltage source, ie, it produces a fixed time-varying potential difference across its terminals irrespective of other components in the circuit. Since the voltage across the AC source is independent, the voltage source across the rectifier is also independent.


Correct.


  Now, according to Kirchoff's laws, the potential difference across the terminals of the rectifier must be equal to the potential difference across the capacitor.


Not correct. There can be a voltage across the diodes if they're off (not conducting). To see this more clearly, imagine removing the load and leaving only the capacitor and the diode rectifier. As the AC voltage rises, it charges the capacitor through the diodes. But when the AC voltage starts to fall again, it can't pull charge out of the capacitor -- the diodes only conduct one way! (That's the whole point of a rectifier.) The capacitor is still charged to the max AC voltage and stays that way forever. The AC source never supplies any more current.

Back to the real world. With a load, the capacitor drains over time. At the peak of the AC half-cycle, the AC voltage becomes greater than the capacitor voltage. The diodes turn on and the AC source charges the capacitor back to its maximum value. This is shown at the bottom of your picture.See the dashed purple line in your picture labeled "Waveform without capacitor"? That's the input voltage from the AC source. The black line labeled "Waveform with capacitor" shows the capacitor being charged up at the peak of the half-cycle, then draining slowly due to the load once the diodes turn off.

A higher capacitance means the capacitor (output) voltage drains more slowly:

$$\frac {dV_{Capacitor}} {dt} = \frac {I_{Load}} {C}$$

UPDATE: You asked whether the voltage across the rectifier is equal to the voltage across the load. The answer is yes, but only when the diodes are on. In that case, one pair of diodes acts (almost) like a short circuit, and you get this:







simulate this circuit &ndash; Schematic created using CircuitLab

When the diodes are off \$(V_{Cap} &gt; V_{AC})\$, they act like open circuits, and you get this:







simulate this circuit

Diodes are nonlinear circuit elements. Their behavior changes based on the voltages and currents in the rest of the circuit!

For KVL, you need to look at it like this (simplified a bit):







simulate this circuit

One possible loop equation is:

$$V_{AC} = V_{D1+D2} + V_{Load}$$

Or:

$$V_{AC} = V_{D1+D2} + V_{Cap}$$

since the capacitor and load are in parallel. Now when the diodes are off, the cap/load is completely disconnected from the AC source ("floating"), so figuring out the diode voltage gets a little weird, especially since we're using ideal diodes. Fortunately, you don't have to worry about it. When the diodes are off, you can use KVL on the capacitor/load loop:

$$V_{Cap} = V_{Load}$$

(Of course, real diodes have a forward voltage drop when they conduct, but I'm leaving that out here for simplicity. You can put it in the \$V_{D1+D2}\$ term in the KVL equations above.)