At what voltage does the electrodeposition of the metal start?
We can't write a relation for V, V1 and V2, because we can apply virtually any V, and still have V1 and V2 constant :) Though, if V signifies the minimum applied potential for starting reactions, then it's ok.
Instead, V may be compared to total potential of cell, which is V2-V1 or V1-V2 (see [3] below about the sign), but not V1+V2, as you assumed.
From wikipedia:
E°cell = E°cathode − E°anode.
The exact sign of (V1-V2) is hard to remember and a subject to mess, so I would better remember the processes involved:
positive voltage means lack if electrons, i.e. oxidation. But (citation)
since the electrode potentials are conventionally defined as reduction potentials (literally: ability to be reduced), the sign of the potential for the metal electrode being oxidized must be reversed when calculating the overall cell potential.
In order to have a spontaneous reaction (ΔG° &lt; 0), E°cell must be positive (see link above). The gold would deposit on the cathode spontaneously if you short external circuit and provide conditions for anode reaction with lower E°. This may be oxidation of OH- (alkaline solution) to O2 on inert electrode. Why do you wonder? Au3+ is powerful oxidizer.
Other note: for high-ionic mix (1M + 1M + 1M + 1M + anions, probably such a mixture is not possible at all) the real redox kinetic will easily be different from simple potential calculation because of polarization effects and other factors. For example, 1-molar Au3+ is probably stable only at low pH, which shifts redox potentials.