555 Timer circuit analysis
What is V at the time the circuit is connected to supply? To me, it seems like it's floating without a reference voltage, since C3 is empty (i.e. basically open circuit) and D2 does not conduct since OUT is low. (I am pretty sure I am wrong here, but I don't know why/where).
  


V is 0 volts. And no, it isn't open-circuit. A capacitor will "hold" 0V just as well as it holds any other voltage.


  
  How to determine the period and the duty time? (I am looking more for a formula and some pointers to understanding its derivation)
  


In astable mode, V is cycling between 1/3 Vcc and 2/3 Vcc. It does this at a rate determined by both the capacitance and resistance that is applied during each phase of operation.

It can be shown that the exponential curve moves from 1/3 to 2/3 (or vice-versa) in a time that is equal to \$-\ln (0.5) \cdot R \cdot C\$, or 0.693&times;R&times;C, which is the formula used in the datasheet.

EDIT: Specifically, the voltage across the capacitor (during discharge) is
$$v(t) = V_0 e^{-\frac{t}{RC}}$$
We want to know how long it takes to fall from 2/3 Vcc (\$V_0\$) to 1/3 Vcc. Therefore, we set the expression equal to 1/3 Vcc and solve for t:
$$\frac{1}{3}V_{CC} = \frac{2}{3}V_{CC} \cdot e^{-\frac{t}{RC}}$$
Isolate the exponential:
$$\frac{1}{2} = e^{-\frac{t}{RC}}$$
Take the logarithm of both sides:
$$\ln \frac{1}{2} = -\frac{t}{RC}$$
And finally, isolate t:
$$-\ln \frac{1}{2}\cdot R \cdot C = t$$
A similar formula applies when charging from 1/3 Vcc to 2/3 Vcc using a supply voltage of Vcc.


  
  The examples of 555-based timers I have seen around usually charge C3 from Vcc. In this case, it is charged from the output pin. Are there any advantages/disadvantages of this approach?
  


This circuit topology allows the timer to operate with a duty cycle that is less than 50% &mdash; in this case, it can be MUCH less than 50%. A disadvantage is that the charging voltage is not exactly Vcc (and this gets worse if there's a significant load on the output), which throws off the timing calculation for the "on" time.


  
  What is the role of D1?
  


The relay is an inductive load, which means that there's a significant voltage spike when you turn it off that could damage the output of the IC. D1 "shorts out" this spike to protect the rest of the circuit.