It's put in a bit an unclear way. It's all about Lorentz-transformation properties of the electromagnetic field and the Lorentz force.
The situation discussed is the following: In the inertial reference frame A you have a wire with a DC current density ##\vec{j}## producing a magnetic field. Neglecting the very small relativistic effects (or the Hall effect for the conduction electrons) in this reference frame the charge density is ##\rho=0##. Then you have no electric field ##\vec{E}=0## but a magnetic field ##\vec{B}=B \vec{e}_{\varphi}##, where I use cylinder coordinates ##(\rho,\varphi,z)## with the wire along the ##z## axis.
A charged particle moving with a velocity ##\vec{v}=v \vec{e}_z## along the wire feels a purely magnetic Lorentz force,
$$m c \mathrm{d}_{\tau} \vec{u}=q \vec{u} \times \vec{B}.$$
Here ##\tau## is the proper time of the charged particle and ##\vec{u}=\gamma \vec{v}## are the spatial components of the four-velocity
$$u^{\mu} = \frac{1}{c} \mathrm{d}_{\tau} x^{\mu}.$$
The forth component of the equation of motion follows from ##u_{\mu} u^{\mu}=1=\text{const}##, which implies
$$u_{\mu} \mathrm{d}_{\tau} u^{\mu}=0 \; \Rightarrow \; u_0 \mathrm{d}_{\tau} u^0=\vec{u} \cdot \mathrm{d}_{\tau} \vec{u}=0$$
in our case, i.e., the 4D invariant Minkowski force on the particle at the initial time is
$$K^{\mu}=\begin{pmatrix} 0 \\ q \vec{u} \times \vec{B} \end{pmatrix}=\begin{pmatrix}0 \\ -\frac{q B v \gamma}{c} \vec{e}_{\rho} \end{pmatrix}.$$
Here we've used the usual notation ##\gamma=1/\sqrt{1-v^2/c^2}=u^0##.
Now consider the situation in another inertial frame B, where the particle is initially at rest. This is described by a Lorentz boost in ##z## direction with boost velocity ##v##.
The current density together with the charge density makes up a four-vector field. In frame A it has the components
$$(j^{\mu})=\begin{pmatrix} 0 \\ 0 \\ 0 \\ j \end{pmatrix},$$
and in frame B, according to the Lorentz transformation the components
$$(\tilde{j}^{\mu})=\begin{pmatrix} -\beta \gamma j \\ 0 \\ 0\\ \gamma j \end{pmatrix}.$$
Since in this frame the charged particle is at rest at the moment it moves parallel to the wire in A it only fields the electric force. Since the wire is charged now, in this frame is also an electric field, which is also given by the Lorentz transformation (the electromagnetic field is represented as an antisymmetric four-tensor field in relatitivity which determines its transformation properties). In our case we have
$$\vec{\tilde{E}}=\frac{\gamma}{c} \vec{v} \times \vec{B}=\frac{\gamma v B}{c} \vec{e}_z \times \vec{e}_{\varphi}=-\frac{\gamma v B}{c} \vec{e}_{\varrho}.$$
The Minkowski force on the particle at that moment as measured in frame B thus is
$$\vec{\tilde{K}}=q \vec{\tilde{E}} = -\frac{q\gamma v B}{c} \vec{e}_{\varrho}.$$
The temporal component follows again from
$$\tilde{u}_0 \mathrm{d}_{\tilde{t}} \tilde{u}^0=\mathrm{d}_{\tilde{t}} \tilde{u}^0 = \vec{u} \cdot \mathrm{d}_{\tilde{t}} \vec{u}=0.$$
This you get also directly from the Lorentz boost of ##K^{\mu}##.
Note that for the usual non-covariant Force ##\vec{F}=\vec{K}/u^0## there's a ##\gamma## factor
$$\tilde{\vec{F}}=\vec{\tilde{K}}{\tilde{u}^0}=\vec{\tilde{K}}=\gamma \vec{F}.$$
The relation of all this to Lorentz contraction is a bit indirect. You can derive everything from the Lorentz transformation properties of the charge-current-density four vector. In frame A the positively charged ions making up the wire are at rest and the conduction electrons are moving. In our approximation, neglecting the Hall effect, in this frame the total charge density (positive ions + conduction electrons) is set to 0. Now in frame B the positively charged ions are moving and thus their charge density gets a ##\gamma## factor compared to frame A because of the length contraction of the volume elements used to define the charge density. The conduction electrons are moving in frame A and thus their charge density does not only get a Lorentz factor in frame B but there's also a component from the spatial components of the corresponding current density, and thus the charge density of the conduction electrons does not compensate precisely the charge density of the positive ions when measured in frame B. The net result is the above given negative charge density of the wire as a whole, which implies the existence of the radial electric field, which is the cause of the purely electric Lorentz force in frame B.
A very nice discussion of the relativistic formulation of electromagnetism (which simplifies a lot compared to the usual non-relativistic heuristics used in conventional textbooks) can be found in
M. Schwartz, Principles of Electrodynamics, Dover (1987)
or
L. D. Landau, Course of Theoretical Physics, vol. 2, Butterworth-Heinemann (1996)