I entered this into LTSpice to help visualize what happens, which I highly recommend.
At DC with the switch S1 open, R3 provides base bias to T2 turning it on. LED2 stays lit. At this point T2 is saturated (collector ~= 0.2V). R5 holds the base of T1 at about 0.2V, turning it off.
The left side of C1 is around 8V and the right side is at approx 0.6V. So far so good.
When the user closes switch S1, T1 turns on grounding the left side of C1. This will cause the right side of C1 to transition DOWN by about 8V to somewhere around -7V. T2 will see negative Vbe of 7V which WILL be an issue for a BC548. The AbsMax VEBO rating is 5V and the circuit applies 7V. Not good.
Now the right side of C1 begins charging through R3. This is the monostable pulse portion. The right side of C1 slowly charges until reaching about 0.6V and T2 turns on. The collector of T2 goes back to 0.2V and R5 causes Q1 to turn off. This terminates the pulse.
In the picture above:
Blue trace: Switch S1 (high = switch closed)
Red trace: Current through LED1 (the pulse)
Green trace: Voltage across the capacitor
It's true that C1 does become reverse biased by about 0.5V at the end of the charging cycle. The max reverse bias can be calculated as T2_Vbe - T1_Vce_sat. Use a non-polarized cap if you want to avoid trouble.