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
$$
v_g=\frac{\Delta\omega}{\Delta k}
$$
With a continuous wave you cannot transmit a signal. For a signal to be transmitted, you need a modulation of the wave, e.g. amplitude modulation. For example, to transmit acoustic frequencies (speech), you modulate the high frequency electromagnetic carrier wave (on the order of MHz for medium wave transmitters) with the acoustic frequencies(up to 20kHz). This modulation produces small variations called side-bands (plus and minus 20kHz) in the transmitted waves. The group velocity of a wave describes the velocity with which such modulation of the carrier amplitude, which transmits the signal, propagates. In free space, the group velocity of an EM wave is identical to the phase velocity $c$ because the dispersion is linear $\omega=c k$. Thus also a pulse shaped modulation propagates with unchanged form. On transmission lines, there can be significant nonlinear dispersion, i.e. the phase velocity $v_{ph}= \frac {\omega}{k}$ for different frequencies is not constant and, in general, different from the group velocity $v_{gr}=\frac {\partial \omega}{\partial k}$. This leads to a loss of shape of a pulse-like modulation of the carrier wave. However, the propagation speed of such a pulse modulation can still be obtained from the group velocity.
That the group velocity is opposite to the phase velocity happens only in systems with special nonlinear dispersion relations.
Best Answer
Consider a wave
$$A = \int_{-\infty}^{\infty} a(k) e^{i(kx-\omega t)} \ dk,$$
where $a(k)$ is the amplitude of the kth wavenumber, and $\omega=\omega(k)$ is the frequency, related to $k$ via a dispersion relation. Note, if we wanted to track a wave, with wavenumber $k$, with constant phase, we would see that this occurs when $kx=\omega t$, i.e. $x/t = \omega/k = c$, with $c$ the $\textbf{phase}$ velocity.
We would like to know the speed at which the envelope $|A|$ is traveling.
For $\textbf{narrow banded}$ waves, the angular frequency $\omega$ can be approximated via the taylor expansion around a central wavenumber $k_o$, i.e.
$$\omega(k) = \omega(k_o) + \frac{\partial \omega}{\partial k} (k-k_o) + \mathcal{O}((k-k_o)^2),$$
where the scale of the bandwidth is quantified by the small parameter $(k-k_o)$. Therefore, we can rewrite $A$ as
$$A \approx e^{-i(\omega(k_o)t-k_o\frac{\partial \omega}{\partial k}t)} \int_{-\infty}^{\infty} a(k) e^{ik(x-\frac{\partial \omega}{\partial k} t)} \ dk.$$
Therefore
$$|A| = \left| \int_{-\infty}^{\infty} a(k) e^{ik(x-\frac{\partial \omega}{\partial k} t)} \ dk \right|,$$
which says that the envelope, $|A|$, travels at speed $\frac{\partial \omega}{\partial k}$, i.e. $$|A(x,t)| = |A(x-c_g t,0)|,$$ where we have defined $$c_g \equiv \frac{\partial \omega}{\partial k}.$$
The group velocity has dynamical significance, as it is the velocity at which the energy travels.