Units of a Fourier Transform (FFT) when doing Spectral Analysis of a Signal

take a signal, a time-varying voltage v(t)

units are V, values are real.

throw it into an FFT -- ok, you get back a sequence of complex numbers

units are still V, values are complex ( not V/Hz - the FFT a DC signal becomes a point at the DC level, not an dirac delta function zooming off to infinity )

now take the modulus (abs)

units are still V, values are real - magnitude of signal components

and square the result, i.e. |fft(v)|^2

units are now V², values are real - square of magnitudes of signal components

shall I call these spectral coefficients?

It's closer to an power density rather than usual use of spectral coefficient. If your sink is a perfect resistor, it will be power, but if your sink is frequency dependent it's "the square of the magnitude of the FFT of the input voltage".

AT THIS POINT, you have a frequency spectrum g(w): frequency on the x axis, and... WHAT PHYSICAL UNITS on the y axis?

Units are V²

The other reason the units matter is that the spectral coefficients can be tiny and enormous, so I want to use a dB scale to represent them. But to do that, I have to make a choice: do I use the 20log10 dB conversion (corresponding to a field measurement, like voltage)? Or do I use the 10log10 dB conversion (corresponding to an energy measurement, like power)?

You've already squared the voltage values, giving equivalent power into a perfect 1 Ohm resistor, so use 10log10.

log(x²) is 2 log(x), so 20log10 |fft(v)| = 10log10 ( |fft(v)|²), so alternatively if you did not square the values you could use 20log10.

The y axis is complex (as opposed to real). The magnitude is the amplitude of the original signal in whatever units your original samples were in. The angle is the phase of that frequency component.