But the row of moving electrons is contracted in the lab frame, compared to what the cat sees. You can see that in your screencaps from the video too -- Derek sees 10 electrons per image width, whereas in the rest frame of the electrons there are only about 8½ electrons per image width.
What is potentially confusing is that as far as the electrons themselves are aware (electrons are not "aware" of anything, but never mind), they are not that the same mutual distance when they are moving as when the wire carried no current.
In other words, the row of electrons is not a rigid object. If each pair of neighboring electrons had been separated by a little rigid rod, the electrons would have to come closer together when the current starts flowing. But there are no such rods, and the row of electrons is free to stretch when the current begins to flow, and this stretching is exactly canceled out by the length contraction, such that in the lab it looks like the distance between the moving electrons is the same as the distance between the stationary protons.
What does the cat see? When the wire didn't carry current, the electrons and protons moved backwards together at the same speed (and with the same contracted spacing as the cat sees the protons have during the entire experiment). Then, when the current starts to flow, the electrons in front of the cat begin coming to a halt (with respect to the cat) slightly before those behind it. So from the cat's point of view the row of electrons get significantly stretched.
Meanwhile, Derek will see all the electrons begin to move at the same time. The cat and Derek do not agree whether two electrons changed their velocity at the same time or not -- this is relativity of simultaneity and is mathematically necessary to make length contraction consistent.
I would leave special relativity out of the picture. I mean: magnetism is a relativistic effect, in the sense that it pops out from the application of (special) relativity to electrostatics, but their relationship is more conceptual and may deserve a new question entirely devoted to the matter.
Long story short, you don't need special relativity to understand magnetic forces.
All you need to know is that when a charged particle $q$ moves at a velocity $\mathbf{v}$ through an electric field $\mathbf{E}$ and a magnetic field $\mathbf{B}$, it experiences the Lorentz force
$$ \mathbf{F_L} = q( \mathbf{E} + \mathbf{v}\times\mathbf{B}) $$
This really is the fundamental equation for electrodynamics, together with Maxwell's equations.
Now, a current-carrying wire is usually considered to be neutral overall (i.e. the number of electrons and ions are the same so that the net charge is zero), so you can take the electric field $\mathbf{E} = 0$.
Then, you are dealing with a current, not with a single charge. How to get the above equation in terms of the current $I$?
The electric current $I$ is defined to be $\frac{dq}{dt}$ where $q$ is the electric charge. The velocity is defined as $\frac{d\mathbf{l}}{dt}$ where $d\mathbf{l}$ is the path element in the direction of the current (or of the charge's trajectory).
So, the infinitesimal force on an infinitesimal charge $dq$ is $d\mathbf{F_L} = dq( \mathbf{v}\times\mathbf{B}) $, where
$$ dq\cdot\mathbf{v} = dq\cdot\frac{d\mathbf{l}}{dt} = \frac{dq}{dt} d\mathbf{l} = Id\mathbf{l}$$.
To get the total force, you just integrate along the path defined by $d\mathbf{l}$:
$$\mathbf{F_L} = \int_L I d\mathbf{l}\times\mathbf{B} $$
OK. But you have two wires.
And you know that a current carrying wire generates a magnetic field, given by (assuming the wire is in a straigh line) :
$$ B_{\phi} = \frac{\mu_0}{2\pi}\frac{I}{r} $$
meaning that it is only in the tangential ($\neq$ radial) direction. The direction of the field is given by the right hand rule, line up your right thumb with the current direction and your fingers will tell you the direction of the magnetic field. This formula is obtained via Ampère's law.
If the two wires are in a straight line and are separated by a distance $d$, the value of the magnetic field due to one at the position of the other is $B = \frac{\mu_0}{2\pi}\frac{I}{d}$. Assume the wires carry equal, constant currents $I$.
Using this as the magnetic field in the Lorentz force, you get a force of
$$ |\mathbf{F}| = I\cdot(\frac{\mu_0}{2\pi}\frac{I}{d})\cdot \int_L dl = \frac{\mu_0}{2\pi}\frac{I^2}{d}\cdot L $$ where $L$ is the length of the wires, or a force per unit length of $$ |\mathbf{F}| = \frac{\mu_0}{2\pi}\frac{I^2}{d} $$.
$d\mathbf{l}$ points along the current: use this and the cross product in the Lorentz force equation to get the direction of the resulting force.
If the currents are parallel, the force is attractive, while if they are opposite, the force is repulsive. The magnitudes of the forces are the same, as they only depend on the wire separation $d$ and on the current $I$.
Notice, the other wire will exert the exact same force on the fist, due to Newton's III law of motion.
Best Answer
As far as your comment goes, you mean there is an absolute symmetry between the 2 wires. Maybe, but one thing I must tell you that when you are considering the electrons in WIRE 1, the relativistic effects will be as follows:
Hope your doubt has been resolved.