Reducing Ripple in the Zener Regulator
note: This answer also addresses some issues mentioned in the comments of the question, have a look there too.

I'll redraw the schematic for you:







simulate this circuit &ndash; Schematic created using CircuitLab

Note that I've added \$R_S\$ ('s' as in source). Let's analyze the circuit starting from the output.

Q1 is in common collector configuration, meaning that its voltage gain is near unity while current gain is much higer, given that it's in active region. The voltage across \$R_{load}\$ is approximately the base voltage diminished by a \$V_{BE}\approx0.7\text{V}\$.

Assuming that DZ1 is working properly the base voltage is set by it. The rightmost R must provide enough current for the diode and for the BJT though.

The diode and transistor bias current is drawn from the LP filter output. The LP filter is a classic RC first order filter, note that its output sees a load approximately equal to R (DZ1 is a short circuit for small signals).

The LP filter is powered by the PSU, that also feeds the bjt collector through \$R_C\$.

Now to your questions:

Why is the leftmost R needed?
That's because \$R_s\$ is very small. Using only a capacitor would result in an LP filter but its corner frequency would be too high. A source output resistance is not something you want to rely on anyway, that's probably not well characterized.

Can we use a diode instead of the leftmost R? If yes, why?
You can use a diode instead of R effectively building a peak detector.

What is better? Diode or resistor?
Honestly, I am not sure. I guess the answer lies mainly in the PSU spec: keep in mind that a diode there would not limit the current going into the capacitor, that might be a problem. A diode would be slower to follow descending peaks of the PSU but that should not actually be an issue.

Why voltage drop across \$R_c\$ should be less than voltage drop across rightmost R?
That's to keep the transistor in active region. If the collector voltage goes too low (high collector currents -> high drop across \$R_C\$) the transistor may saturate and stop working properly. I think the circuit can work without \$R_C\$.