Trouble understanding a given circuit for an LDR pulse sensor
I assume the transistors are there to amplify the signal variation of the LDR. How do they do that?

To a first approximation, think of the transistor as having two current flows, one from base to emitter, the other from collector to emitter. (Yes, I know that sounds backwards, but in electronics, instead of thinking about electron flow, we think of flow as going from positive to negative. Just accept that. :-) ). Anyway, the collector current (simplified) is controlled by the base current. The ratio between these currents is called \$\beta\$, and a value of 100 is a good guess.
One other fact you need to know before you can apply Kirchoff's laws. There is a voltage drop of somewhere around 0.65V between the base of the transistor and the emitter. The exact value changes with temperature and current, but if you use 0.6-0.7 you will be in the ball park. You need to know this because there are resistances in parallel with the base-emitter path in your circuit.
From there, you may be able to apply Kirchoff's laws. Try two different values for the resistance of your LDR and see what that gives you.

From what I understood from my research about these transistors, there is a 0.7V drop between base and emitter, and base current is multiplied by a constant to determine collector current.

Yes

I tried using these to calculate an expression for the voltage measured in the analog pin in function of the LDR resistance, to understand the effect, but I did not get anywhere decent.

Try first solving for just the battery, LDR, R3 and the base-emitter path of Q1.



simulate this circuit &ndash; Schematic created using CircuitLab
Then multiply the base current you get by \$\beta\$ (approx. 100) and see if you can solve the rest.