Why do I need two PNP BJTs as a switch rather than only one?
Assuming that the part number and package of the component U1 are correct, the pin 14 is the supply voltage pin of the PIC. The bipolar-based circuit is essentially connecting or disconnecting the PIC to the VCC.
Now to the circuit itself, it works as follows:
If the switch \$SW_1\$ is open

The bipolar \$Q_1\$ is off because its emitter and base voltages are the same (diode is not forward biased)
Since \$R_2\$ and \$R_3\$ are pulling the base of \$Q_2\$ to ground, \$Q_2\$ will be fully on, and will consequently bias the PIC (I could not figure out the reason for the resistor \$R_4\$

Now if the switch \$SW_1\$ is closed

The emitter-base diode of the transistor \$Q_1\$ will turn on and consequently it will pull the node between \$R_2\$ and \$R_3\$ to appoximately VCC, thus turning off the transistor \$Q_2\$
Now the PIC is off

Note: The use of PNP and NPN bipolars depends on the application. Generally PNPs are used as high side switches. The switch acts as a source (current flows out of the switch and into the load). NPNs are generally used as low side switch. The switch acts a sink for the load (current flows from the load and into the switch).
UPDATE#1
As for the further clarifications:
Jumper \$J_5\$
The output pin \$RTS\$ of the FTDI chip has a logic level high (4.1V) when its input buffer is full, and a low logic level (0.4V) when it can still accept data. I reckon the reason behind placing a jumper there, is to add a software based reset driven by the FTDI chip. The way the circuit is wired right now, it would work as follows if the jumper would be connected and \$SW1\$ is open.
\$RTS\$ is low

FTDI is going to pull the base of \$Q_1\$ low, thus turning it on and consequently turning \$Q_2\$ off, and resetting the PIC.

\$RTS\$ is high

Nothing much really changes since the FTDI would still try to pull up the base voltage of \$Q_1\$ to, thus turning it off. In this scenario the PIC is running and probably

The reason why the designer opted for a jumper instead of directly connecting the \$RTS\$ pin might be that, in case the \$RTS\$ pin was high (4.1V) and the switch \$SW_1\$ is pressed at the same time, you would short circuit the output pin to ground. Since this pin can source maximum 2mA, it could probably damage it.
Capacitor \$C_3\$
It simply connects the supply pin of the PIC to ground and acts as a bypass capacitor, handling high current transients.