In the original formula we have that $$\omega_1t-k_1x = A$$say, and $$\omega_2t-k_2x = B$$say, by hypothesis for a specific point at time $t$ and position $x$. This is a point of constant phase (for the $A$ wave and $B$ wave respectively.) To determine the velocity of a (sine or cosine) wave from first principles one wants to know the velocity of that point: how far does a point of constant phase move in time t? This gives the answer of the phase velocity
$$
v_p=\frac{\omega}{k}
$$
So the component waves are moving with phase velocities: $v_{p1}$ and $v_{p2}$ respectively. Using the above values for $A$ and $B$ the middle derivation is an application of the trigonometric identity:
$$
sin A + sin B = 2 sin (1/2(A+B)) \cdot cos (1/2(A-B)).
$$
This gives your expression for $s = s(x,t)$. So how can one talk about a point of constant phase here to obtain the group velocity as it is a product of sin and cos?
Well the trick indeed is to recognise the two separate wave components and treat these (for now) as two separate waves and calculate their (phase) velocity - ie the rate of movement of points of constant phase in each "wave".
For the sine wave ie the envelope we would get $\frac{\overline{\omega}}{\overline{k}}$.
Now for the cosine wave. The short answer is that we are looking for its phase velocity also, namely $\frac{\Delta{\omega}}{\Delta k}$.
However what your professor has done here, is to calculate from first principles the velocity of that cosine wave. That is to ask for the definition of a point of constant phase, viz: $\frac{\Delta{\omega}}{2}t - \frac{\Delta k}{2} x = const$ and then to determine the velocity (by differentiation, etc) of this point, again resulting in
$$
v_g=\frac{\Delta\omega}{\Delta k}
$$
I think the wave equation can by derived from geometry alone, without using physics. Consider $f(x-ct)$ and consider small changes in $x$ and $t$, ie. $\Delta x$, $\Delta t$ (They each cause a small shift or translation of $f(x-ct)$). Note that $\Delta x$ = $c\Delta t$. So $\frac{\Delta f}{\Delta x}$ = $\frac{\Delta f}{c\Delta t}$ = $\frac{1}{c}\frac{\Delta f}{\Delta t}$. Doing that again we get
$$\frac{\Delta^2 f}{\Delta^2 x} = \left(\frac{1}{c}\right)^2\frac{\Delta^2 f}{\Delta^2 t}$$. Then letting $\Delta$ become very small we get
$$
\frac{\partial^2f}{\partial x^2}-\frac{1}{c^2}\frac{\partial^2f}{\partial t^2}=0
$$
From the geometry alone, it was only needed to note that a change in $t$ multiplied by the velocity yields the same results (as measured by the second derivative) as a change in $x$ --that is, a translation of $f(x-ct)$. See, for example, for a good discussion: kiskis.physics.ucdavis.edu/landau/phy9hc_03/wave.pdf.
Best Answer
Notes: I believe that the answer you are trying to obtain is incorrect. First, if you are looking at total energy density, you should not have the factor of 1/2. Alternatively, you could be looking at either the potential or the kinetic energy density, not the sum. Second, I believe that $s$ in your solution should be $s_0\cos(kx-\omega t)$. Alternatively, you could be referring to the mean total energy density, in which case $s$ should be replaced with $s_0$, and the factor of 1/2 is now accurate.
Another note is that the expression for $s$ you have is a traveling wave. The analysis is different for standing waves, though it follows the same steps.
Derivation:
We will assume, as you have in your problem, that the displacement of the sound wave is given by $$s = s_0\sin(kx-\omega t).$$ Then the particle velocity may be obtained by taking the time derivative: $$v = -\omega s_0\cos(kx-\omega t).$$ We may then use the continuity equation to obtain an expression for the pressure. The continuity equation is given by $$\frac{1}{c^2}\frac{\partial p}{\partial t} + \rho\frac{\partial v}{\partial x} = 0,$$ where $c$ is the speed of sound and $\rho$ is the ambient mass density, which leads to the solution $$p = -\rho c^2ks_0\cos(kx-\omega t).$$
Now consider a single fluid element of volume $dV$. The mass of this element is $\rho dV$. The total kinetic energy of this element is then given by $$dE_\text{k}=\frac{1}{2}\rho dV v^2 = \frac{1}{2}\rho\omega^2 s^2_0\cos^2(kx-\omega t)\ dV.$$ The potential energy comes from the fact that the fluid element acts like a spring, and so may be written as one half of the force times the compression. The force is the pressure times the cross-sectional area $dA$, and the compression is minus the derivative of the displacement times the element length $dx$. Noting that $dx\ dA=dV$, we may then write $$dE_\text{p} = -\frac{1}{2}p\frac{\partial s}{\partial x}dV = \frac{1}{2}\rho c^2k^2s_0^2\cos^2(kx-\omega t)\ dV.$$ By definition, we know that $k^2=\omega^2/c^2$, and so we may then write $$dE_\text{p} = \frac{1}{2}\rho \omega^2s_0^2\cos^2(kx-\omega t)\ dV.$$
If we divide the kinetic and potential energies by $dV$, we obtain expressions for the respective energy densities. The total energy density is then given by $$\frac{dE_\text{k}}{dV} + \frac{dE_\text{p}}{dV} = \rho\omega^2s_0^2\cos^2(kx-\omega t).$$
We may be interested in the mean energy density, in which case we integrate the energy density in time over one period and divide by one period. This process leads to $$\frac{\omega}{2\pi}\int_0^{2\pi/\omega}\left[\frac{dE_\text{k}}{dV} + \frac{dE_\text{p}}{dV}\right]dt = \frac{\rho\omega^3s_0^2}{2\pi}\int_0^{2\pi/\omega}\left[\frac{1 + \cos(2kx-2\omega t)}{2}\right]dt = \frac{1}{2}\rho\omega^2s_0^2.$$