Many of the things you write sound OK. But I wouldn't say that the other directions in the graph are mere strengths, they indicate the actual value (strength/magnitude and direction as well) of the electric and magnetic fields.
First note that technically the electric field is a vector and it is a field so it should have a vector (possibly zero at every point). Each vector has a head and a tail. You can think of the location of the tail as telling you the place where the field has a particular value. Then you can think if the difference between the head and the tail as telling you both the magnitude and the direction of the electric field at that point, in some specified unit system. So imagine a bunch of arrows all the same color the location of each is telling you where it is telling you the field and how the arrow points from there tells you the value.
Then draw the magnetic fields in a different color. And in both cases you can't draw a vector at every single point because it would just be too much to see them all.
For a plane wave traveling through vacuum there is a great deal of regularity.
If the wave travels in the x direction then the electric and magnetic fields all point in the y-z plane. And their values only depend on time and on x, so you can draw just one longitudinal line and it will tell you about all of the values at all the locations. And since the electric fields and magnetic fields all point in the y-z plane you can reimagine the y-z plane as like the independent axis of a graph.
And for a classical plane wave travelling through a vacuum your picture should have the electric and magnetic fields both be strong together at one plane of a fixed value in the longitudinal direction and then at a different plane corresponding to a different fixed value of the longitudinal direction. Strong together and weak together and then strong again but pointing in the opposite direction.
If so, you get a picture much like the first one you drew (though that is just for one the electric and magnetic fields, a wave has both). That is all fine for a classical wave travelling through a vacuum.
For your other questions I think you should look at existing questions about the different topics.
From a quantum perspective it is much more complicated. From a quantum view there isn't an electric or a magnetic field, there is a photon field. When you have a large number of photons all in phase with each other it can look or act like an electromagnetic wave, but it is still different, and if you have a small number or they aren't in phase then it just truly different.
The electric and magnetic fields classically are stand-ins for saying how charges interact (though they do have their own energy and momentum, pressure and stress, etc.) And in quantum mechanics different interactions are possible and so the photon field is a stand in for saying how those different interactions go.
Since in quantum mechanics the charges don't have a location and momentum saying that their momentum changes based on the field where the charge is located just isn't going to be possible because you have none of those things. And if you can't verify the field's values at any location it becomes difficult to say it is there in that way.
We make the objects we need to predict results, for quantum mechanics we need different objects because we are predicting different interactions. Only in some limits (that only hold sometimes) do we expect quantum mechanical effects to start to approximately look like classical effects.
As for a transverse spacing in a Faraday cage stopping a wave that varies only in the longitudinal direction remember that you were drawing only the simplest case, the case of a plane wave.
In a plane wave the entire plane parallel to the y-z plane has the exact same electromagnetic field. This very simple (to describe) solution allows the wave to travel entirely in the x direction, over time each part of the wave simply slides over, the new value at distance $x=x_0+\Delta x$ over is the old value at $x_0$ a time $\Delta x/c$ earlier.
But no real wave is ever that perfect. They might expand in a spherical front instead of a plane front. They might have a central region where it is strong and it gets weaker farther out (like a Gaussian beam) some people even make beams where the energy and momentum travels along a cone so there is a central strong region but if you place a small object there the region beyond it still gets light because the light from farther out came from a base of the cone that started out farther from the central region.
Sort of like if you had some racers all lined up for a race but instead if running in their own tracks they ran towards the same point that is in front on the center runner. The center runner gets there first, but if parts of the center of the track are damaged the runners that left later will eventually close in and it won't matter that that the runners originally closer to the center got affected.
So a real wave has a more complicated wave front. And in fact if you are trying to avoid the wires of the Faraday cage you need your beam to be focused to not extend out too far in the direction transverse to the direction of propagation. And this isn't restricting the amplitude, this is saying that the actual wave shouldn't have field values spread out in the yz plane, it should be focused to a small portion of the yz plane.
Imagine you shot narrow beams all towards the same hole in the Faraday cage, you need very precision aiming each point has to has "its x" aimed very precisely to get in that same hole.
In general if you aim your initial propagation directions with accuracy on the order of the wavelength then you gave changed the wave.
Thus is because the direction of propagation isn't a magic thing you can freely associate with a point. It depends on how the electromagnetic field varied in space. When you field didn't vary in the y and z directions then the field travelled in the x direction. Now you are trying to get each region to aim slightly differently so you try to adjust the electric and magnetic field here and there to have each part aim just right.
Waves tend to spread, so if you aim some at that one whole, lots will be lost in other directions and what aimed well for one whole is nor going to aim for the other holes, so the Faraday cage succeeds at blocking almost all fields coming at it assuming the cage is big enough to have many many holes.
I think many of these question can be (and have been) asked as separate questions.
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
Radio waves will not travel through the Earth. It's just too dense for that.
Think of it this way: when you take an x-ray, you can see your skeleton in detail in the photograph. That's because when a beam of x-rays goes through you, most flesh allows it to pass through with minimal refraction, but it doesn't go through your bones as easily.
Radio waves have less penetration power than x-rays. The planet is considerably denser than you, and it gets denser the deeper you go.
In fact, not even higher-energy radiation like gamma rays would get through the earth. You'd need neutrinos for that, but then, that's not electromagnetic radiation and no, there is no technology to use those to communicate (yet).
Long-range radio communication will usually work with waves that reflect on the ionosphere, have the waves being sent in a very tight beam (you get more range for the same power this way, when compared with omnidirectional transmission), or use relays. It is very usual, for example, to combine the latter two strategies by using sattelites.