EDIT: In the interest of avoiding spreading misleading information, I have removed the portions of this answer that have been disputed or refuted in the comments and edits on this question. Specifically, the parts about the ACK/Distance shown on the screen at 42:47 and the calculation of the curvature have been removed. The rest of this answer, however, still stands.
TL;DR: They erroneously believed that radio antennae were lasers. The antennae should still be able to connect even on a curved Earth.
The video pretends that the signal leaving the radio antennae is like a laser beam, focused in the line that emanates from transmitter to receiver without diverging. In reality, this isn't even close to true, even for directional radio antennae. Both the transmitted signal and the receiver acceptance get wider farther from the respective antennae, purely due to the diffractive properties of waves. This means that the signal actually propagates in a large ellipsoidal region between the antennae called the Fresnel zone**. The rule of thumb that is used in engineering systems is that as long as at least 60 percent of the Fresnel zone is unobstructed, signal reception should be possible.
The maximum radius $F$ of the Fresnel zone is given in the same Wikipedia article by
$$F=\frac{1}{2}\sqrt{\frac{cD}{f}}\,,$$
where $c=3\times {10}^8 \frac{\mathrm{m}}{\mathrm{s}}$ is the speed of light, $D$ is the propagation distance and $f$ is the frequency. Using $D=14 \, \mathrm{km}$ and $f=5.880 \, \mathrm{GHz},$ we see that $F=13.69 \, \mathrm{m}.$ As you can see, the beam expands massively over such a distance. If you cut out the lower $3.84 \, \mathrm{m}$ of that circle, you would find that the fraction of the beam that is obstructed for obstruction height $h$ from the formula for the area of the cut-out portion given here:
$$\frac{A_{\text{obstructed}}}{A_{\text{whole beam}}}=\frac{F^2\cos^{-1}\left(\frac{F-h}{F}\right)-(F-h)\sqrt{2Fh-h^2}}{\pi F^2}\,.$$
Evaluating this expression for $F=13.69 \, \mathrm{m}$ and $h=3.84 \, \mathrm{m}$ gives you an obstruction fraction of $\frac{A_{\text{obstructed}}}{A_{\text{whole beam}}}=0.085.$
So, even on a curved earth, only 8.5 percent of the beam would be obstructed. This is well within the rule of thumb (which required less than 40 percent obstruction), so the antennae should still be able to connect on a curved Earth.
**In reality, propagation of radio waves between two antennae is complicated, and I'm necessarily skipping over a lot of details here, or else this post would become a textbook. What I refer to as the "Fresnel zone" here is technically the first Fresnel zone, but the distinction is not necessary here.
Best Answer
In principle, the scattering of photons off the gas molecules of the flame can carry away heat in some circumstances (this is for example used in laser cooling). If the transition frequencies of the atoms is close to the laser frequency, photons can carry away momentum.
But your laser is such an open system that any scattering of photons off the hot gas molecules of the flame probably just dissipates the heat in all directions. It makes no sense then, to talk about an ensemble of photons with a temperature. If you were to collect some photons, say by shining them into a high-finesse cavity constructed from mirrors, it could perhaps be possible.