Does it means that if I escape a celestial body following a spiral I can virtually reduce to zero this component?
So, if I could use sort of "mass driver" propulsion on an atmosphere-free body like the Moon I can spare the gravity loss at once? Or I should assume that, whatever I do, this affect only the istantaneous delta-v, not the total amount that remain the same and is proportional to g*Dt (see above)?
Yes, a mass driver on the moon would suffer no gravity losses.
2. Now, for a 3 mins ascent before stage separation, for example, 3 * 60 sec * 9.8 m/s^2 ~ 1750 m/s. This looks too much. Am I wrong? Including the gravity turn should add to this result.