Hi SG,
Welcome to ETO. Which county are you at? If you put it next to 'Location' on your user page, it will display in the box at the left of your posts; that will help us to answer your questions.
Like Ron says, a clever approach you have devised for driving two relays with one wire.
I am a bit confused by your figures though. You say that the relays function with half wave rectified 12V AC at 50Hz (I know that because I know you are from the UK). But then you say that the relays have a resistance of 28K and operate with a current of 4.5mA. That implies a voltage of 126V.
You ask how you calculate the capacitors:
The formula you need to know is Q=CV=IT
Where
Q= charge in coulombs
C= capacitance in Farads
V= voltage in Volts
I= current in Amps
T= time
The time is simple: 50Hz has a period of 20mS. With half wave rectification you have to wait a full period (to a first approximation) before the capacitor gets topped up again. So the discharge time of the capacitor is roughly 20mS. Thus T in the above formula is 20mS.
I, you have stated is 4.5mA
C, you have stated is 1uF
So you need to work out what V (the ripple voltage is or voltage droop between peaks of the AC supply) is.
Extracting and transposing the above formula, you get V = IT/C
Thus, V = (0.0045A * 0.050S) /0.000,001F = 45V
So the capacitors are not doing much.
I would remove the capacitors. Then put a suitable sized resistor of a suitable power rating in series with the input to the circuit (between live and the two switch wipers). Start with 56K and work down until the relays start operating. Then reduce the resistor by a further 20%.
Then go and have a pint of beer.
spec