Ignoring the resistor, when the switch initially closes, the voltage on the cap charges fairly quickly - limited by the power supply and its local decoupling capacitors, the resistance of the switch and any track resistance and inductance. If the switch momentarily went open circuit, the cap would stay largely charged. This means you get a slightly delayed (maybe 10 - 100 us delay) pulse edge and the cap stays charged even if the switch opened and closed a few times. It doesn't jump up and down because the only discharge path for the cap is via the 10k resistor.
In short you get one rising edge with an initial delay of 10 - 100 us after the switch is first pressed. When the switch is released the 10k discharges the cap in about 5 *CR = 5 milli seconds.
There won't be multiple edges generated and that might be important. As for the 2nd circuit I suspect they want the signal "held-up" for a few milliseconds after the drive from the other chips have disappeared - I suspect this is a timing issue or design fix.