I'll offer a mathematical approach, just in case anyone finds it useful. I don't generally
like math that much, but in this case, I do find it intuitive, explaining well the difference between the two.
In full wave rectifier circuits there is something called average dc voltage. Its formula is 0.6366*Vpeak.
It comes from the fact that if you take a sine wave, flip the negative half-waves to the positive side, the average voltage you get over each full wave or cycle is
$$\frac{1}{2 \pi} \int_0^{2 \pi} \left\lvert V_{pk} \sin t \right\rvert dt = V_{pk} ~ \frac{4}{2 \pi} \approx 0.63661977 ~ V_{pk}$$
The integral calculates the area under the curve, and if we divide that area by the width (\$2 \pi\$, one full cycle), we get the average "height", the average DC voltage. Thus, it indeed is the average value of a rectified sine wave.
Why do we need average voltage when there is rms voltage?
If you recall Ohm's law, (instantaneous) power is \$P = V I = I^2 R = \frac{V^2}{R}\$ (for a purely resistive circuit). If you are more interested in the work (or power) you can get out of some sinusoidal voltage, also assuming that the current will be in the same phase as the voltage (so purely resistive load), then you want to look at the average of the squared voltage instead. Of course, units of "voltage squared" don't mean much, so you do a square root afterwards. This is exactly what root-mean-square is! Such an average for a sinusoidal voltage is
$$\sqrt{ \frac{1}{2 \pi} \int_0^{2 \pi} \left\lvert V_{pk} \sin t \right\rvert^2 ~ d t } = \sqrt{\frac{1}{2 \pi} V_{pk}^2 \pi} = \sqrt{\frac{1}{2}} ~ V_{pk} \approx 0.70710678 ~ V_{pk}$$
Thus, average voltage tells you what the corresponding constant DC voltage would be.
RMS voltage tells you what DC voltage would yield the same amount of power, assuming the alternating current is always in the same phase as voltage (which is a sensible assumption, since we're comparing to DC voltage and current).
The same logic expands to
all waveforms, it's not limited to just sinusoidal ones; only the numerical scale factor changes, depending on the shape of the function. Essentially, for any time-dependent voltage signal \$V(t)\$ we have
$$V_\text{AVG} = \frac{1}{T} \int_0^T \left\lvert V(t) \right\rvert ~ d t$$
and
$$V_\text{RMS} = \sqrt{\frac{1}{T} \int_0^T V(t)^2 ~ d t}$$
A good meter that can tell the average does so by repeated (random) sampling, because
$$V_\text{AVG} \approx \frac{1}{N} \sum_{N} V(t)$$
(Even though the rectified sinusoidal varies between zero and only slightly above \$V_\text{AVG}\$, it stays longer at the higher values than in the lower values, giving the expected average.)
Approximating integrals with sums, to any desired accuracy or resolution, is a common mathematical technique. There are rules to when and how it can be done, but when done so, it is absolutely valid, and does not incur any kind of systematic error. Here, the main thing is that the samples are not taken in phase with the signal, so either at much higher frequency than the signal itself, or randomly.
For RMS, there are two approaches. One is mathematical, using repeated (random) sampling, just doing a bit more calculation (squaring and square root)
$$V_\text{RMS} = \sqrt{\frac{1}{N} \sum_{N} V(t)^2}$$
and the other is by measuring the actual average power, by using a (low resistance) resistor that gets heated by the voltage, with the temperature difference carefully measured.
Thus, I'd say your intuition or guesses were on the right track. Average voltage is used when you are concerned about the voltage itself, and not power. RMS voltage is used when you are putting the voltage to work, as it represents the DC voltage that would do the same amount of work (i.e., yield the same power on a purely resistive load; power as in instantaneous watts, or during some interval as watt-seconds or Joules: 1 W s = 1 J.)