1) The potentiometer and the LDR form a voltage divider, which, when it reaches Above 2* 0.65V at the base of the first transistor, it slowly turns on the darlington pair of transistors which will turn on what looks like a relay. The diode is to protect the back EMF from the relay when it turns off.
To summarize, when there is enough light / or not enough, the relay will turn on.
It's a terrible circuit. The darlington will have a really soft turn-on and light changes very slowly. The relay will be pulled in/out unreliably with chattering etc.
For 2 NPN transistors and similar circuit complexity (maybe adding 3 resistors) you can make a schmidt type version of the circuit with some hysteresis that will switch very cleanly.
Ok, I added 2 more resistors and 1 small cap.
The relay transistor is the detector, the other transistor is inverse operation and provides hysteresis via the 1meg resistor.
To reduce the hysteresis use a higher value than 1meg, but I think the 1meg will work ok. It's been a few years since i built a light operated switch like this but those parts values look ok. It should switch clean and fast.
The value of the small cap can be anything, probably 0.1uF or less should be ok.
Note too if you are using a 9v battery and 6v relay you might want to put a resistor (R ?) in series with the relay to keep it's operating voltage at about 6v on the relay coil, this will use less battery power.