If someone asked for the average power dissipated in a device, what would that mean?

The average power is the time average of the instantaneous power.

Consider the case of (in phase) sinusoidal voltage and current.

$$v(t) = V \cos \omega t $$

$$i(t) = I \cos \omega t $$

$$p(t) = v(t) \cdot i(t) = V_m \cos\omega t \cdot I_m \cos\omega t = \dfrac{V_m \cdot I_m}{2}(1 + \cos2\omega t) $$

$$p_{avg} = \dfrac{V_m \cdot I_m}{2}$$

(since the time average of sinusoid over a period is zero.)

In the above, we evaluated the time average of the instantaneous power. In the phasor domain (AC circuits), the product of the rms phasor voltage and current gives the complex power:

$$S = \vec V \cdot \vec I = P + jQ$$

where P, the real part of S, is the average power.

The rms phasor voltage and current for the time domain voltage and current above are:

$$\vec V = \dfrac{V_m}{\sqrt{2}} $$

$$\vec I = \dfrac{I_m}{\sqrt{2}} $$

The complex power is then:

$$S = \dfrac{V_m}{\sqrt{2}}\dfrac{I_m}{\sqrt{2}} = \dfrac{V_m \cdot I_m}{2}$$

Since, in this case, S is purely real, the average power is:

$$P = \dfrac{V_m \cdot I_m}{2}$$

which agrees with the time domain calculation.

If someone asked for the average power dissipated in a device, what would that mean?

The average power is the time average of the instantaneous power. In the case you describe, the instantaneous power is a 1W peak square wave and, as you point out, the average over a period is zero.

But, consider the case of (in phase) sinusoidal voltage and current:

$$v(t) = V \cos \omega t $$

$$i(t) = I \cos \omega t $$

The instantaneous and average power are:

$$p(t) = v(t) \cdot i(t) = V_m \cos\omega t \cdot I_m \cos\omega t = \dfrac{V_m \cdot I_m}{2}(1 + \cos2\omega t) $$

$$p_{avg} = \dfrac{V_m \cdot I_m}{2}$$

(since the time average of sinusoid over a period is zero.)

In the above, we evaluated the time average of the instantaneous power. This will always give the correct result.

You link to the Wiki article on AC power which is analyzed in the phasor domain. Phasor analysis assumes sinusoidal excitation so it would be a mistake to apply the AC power results to your square wave example.

The product of the rms phasor voltage \$\vec V \$ and current \$\vec I \$ gives the complex power S:

$$S = \vec V \cdot \vec I = P + jQ$$

where P, the real part of S, is the average power.

The rms phasor voltage and current for the time domain voltage and current above are:

$$\vec V = \dfrac{V_m}{\sqrt{2}} $$

$$\vec I = \dfrac{I_m}{\sqrt{2}} $$

The complex power is then:

$$S = \dfrac{V_m}{\sqrt{2}}\dfrac{I_m}{\sqrt{2}} = \dfrac{V_m \cdot I_m}{2}$$

Since, in this case, S is purely real, the average power is:

$$P = \dfrac{V_m \cdot I_m}{2}$$

which agrees with the time domain calculation.