= Solution
Use signature $(+,-,-,-)$, which is required for the stated tensor convention. Then $U_b=\gamma(1,-\mathbf v)$. The time component is $F^{0b}U_b=\gamma\mathbf E\cdot\mathbf v$, while $F^{ij}=-\epsilon_{ijk}B_k$ gives the spatial components $\gamma(\mathbf E+\mathbf v\times\mathbf B)$. Thus
$$
\boxed{F^{ab}U_b=\gamma(\mathbf E\cdot\mathbf v,\mathbf E+\mathbf v\times\mathbf B).}
$$
The scalar $\rho_*=J^bU_b$ is the charge density in the medium's rest frame. In that frame $U=(1,0)$, so $J-\rho_*U=(0,\mathbf j)$ and $F U=(0,\mathbf E)$. The rest-frame <Ohm's law> is precisely their proportionality with rest conductivity $\sigma$. Both sides are <four-vectors>, so its covariant form is
$$
\boxed{J^a-(J^bU_b)U^a=\sigma F^{ab}U_b.}
$$
The time and spatial components respectively give
$$
\rho-\gamma\rho_*=\sigma\gamma\mathbf E\cdot\mathbf v,\qquad
\mathbf j-\gamma\rho_*\mathbf v=\sigma\gamma(\mathbf E+\mathbf v\times\mathbf B).
$$
Eliminate $\rho_*$ to obtain
$$
\boxed{\mathbf j=\rho\mathbf v+\sigma\gamma[\mathbf E+\mathbf v\times\mathbf B-(\mathbf E\cdot\mathbf v)\mathbf v].}
$$
If the rest-frame charge density vanishes, set $\rho_*=0$, not necessarily $\rho=0$ in the lab frame. Then
$$
\boxed{\rho=\sigma\gamma\mathbf E\cdot\mathbf v,\qquad
\mathbf j=\sigma\gamma(\mathbf E+\mathbf v\times\mathbf B).}
$$
The induced lab charge is the term that cancels the longitudinal correction in the general formula.
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