Electronic – How does the NPN transistor in a 555 timer discharge the capacitor when the output is low

Use 10kΩ for R3.

I didn't look up a BC547, but if that's a typical small signal transistor as it should be, figure a gain of at least 50. 6V supply minus 700mV B-E drip leave 5.3V accross the resistor. 5.3V / 10kΩ = 530µA. Times the gain of 50, and that can support over 25mA of collector current. That's well more than any reasonable digital input will require, even if it has a pullup connected.

As others have said, you may need a pullup if whatever you are driving doesn't passively make the line go high. However, make sure the pullup goes to same voltage as the power supply of whatever will receive the digital signal. Your supply is 6V, which is probably higher than the receiver's supply. That could cause the receiver to not work or even to damage it if its power voltage is significantly lower.

Okay, looking at the picture I think you may have the transistor the wrong way round.

Try turning it round.

See this picture for reference:

As you can see the collector is on the right with the flat part facing you, so you have the collector connected to the LED in your circuit (if the 2N2222A part you are using has the same pinout)
I got the picture from here.

EDIT - It's actually a 2N222A, but the above advice still goes as the pinout appears to be the same from the picture posted.

As Russell mentions the more standard way is to connect the LED to the collector, but your circuit should work if set up correctly.