Derivation of magnetic force for a charge in motion relative to a wire (M. Purcell's Electricity and Magnetism)

I have a question regarding the derivation of magnetic force in M. Purcell and David Morin's Electricity and Magnetism. In the first part, they derived the force first by assuming that a charge q is moving parallel to a wire and found that there is a magnetic force that acts on the charge, its direction perpendicular to the charge's velocity.

What troubles me is the second part, the one in which they derived the magnetic force that acts on a charge moving perpendicular to the wire. According to the book, in the rest frame of the charge, each two symmetrical electrons (in their distance relative to the charge), which are moving obliquely relative to the wire (or horizontal axis), produce a net electric field parallel to the wire only. They claimed that the net transverse electric field produced by the electrons is utterly cancelled by that of the protons (which are now moving perpendicular to the wire).

What I don't understand is this, and I quote:"The electron on the right always wins. Summing over all the electrons is therefore bound to yield a resultant field E in the $x$ direction. The $y$ component of the electrons’ field will be exactly canceled by the field of the ions. That $Ey$ is zero is guaranteed by Gauss’s law, for the number of charges per unit length of wire is the same as it was in the lab frame. The wire is uncharged in both frames."

How can the wire be uncharged in the charge's rest frame? They argued that the number of protons and electrons per unit length is the same as it was in the lab frame, which is absurd..! The protons' density per unit length in q's rest frame hasn't changed since there is no Lorentz contraction parallel to the wire. However, if the electron's density is indeed still the same in q's frame, as it was in the lab's frame,(which justifies the neutrality of the wire), doesn't that mean that the electrons' velocity component parallel to the wire is still the same? (Else the electron's density per unit length will change)

But as I recall, the electrons' initial velocity (which is parallel to the wire in the lab's frame) should be transformed in q's rest frame (which is moving perpendicular to the electrons and the wire) according to the velocity addition formula! Which means that the electrons density must change and the wire isn't neutral..!

Any help is appreciated! :)

You are correct that the protons' linear charge density doesn't change because any Lorentz contraction happens perpendicular to the wire. You are also right that the relativistic velocity addition formula predicts that the total speed of the electrons isn't simply $\sqrt{v^2+v_0^2}$. It is actually $\sqrt{v_{(\parallel)}^2+\frac{v_{0,(\perp)}^2}{\gamma^2}}$, where parallel and perpendicular refer to the direction of motion of the test charge's frame with respect to the direction of each velocity; the document linked below has a good explanation of the derivation of the total speed formula.
The perpendicular component of the total speed, which is parallel to the wire, is in fact reduced by a factor of $\gamma$. But this doesn't imply that the linear number density of the electrons increases; it only tells you that each electron moves more slowly in the frame of the test charge and that fewer electrons pass in or out of a given segment of wire in a given period of time. This doesn't mean that in any given segment of wire at any snapshot of time there would be fewer total electrons present.