The capacitor is charged when you press the switch.
The capacitor is discharged through the B-E junction of the transistor and RT. The current gain of the transistor means that the current flowing out of the capacitor is a tiny fraction (typically 1/100) of the current flowing through RT. It also means that any bias current required by the 555 inputs is also reduced by a similar factor.
In the "normal" monostable mode of the 555, the capacitor starts out at 0V and charges to 2/3 Vcc, taking 1.1×R×C seconds to do so. In this configuration, the capacitor starts out at Vcc and discharges to 1/3 Vcc plus the VBE drop of the transistor (about 0.65 V), taking somewhat less time to accomplish this.
Note that the sense of the output is reversed: Pin 3 goes low when you push the button, and then goes high again when the time period expires.
This configuration of the transistor is called "emitter follower", and it is used specifically when you want to "buffer" a voltage. The input impedance of this configuration is much higher than the output impedance, and the voltage gain is very nearly unity (albeit with an offset, as noted above).