$U^\dagger (-i U \partial_{k_x}U^\dagger) U = -i (\partial_{k_x}U^\dagger)U$. This expression can be evaluated without an explicit form for $U$. To do this, we first note that by definition $U\hat{b}\cdot\vec{\sigma}U^\dagger = \sigma_z$ where $\hat{b} = \vec{B}/|\vec{B}|$. Thus, $$\begin{aligned}
    && U^\dagger\sigma_zU = \hat{b}\cdot\vec{\sigma} \\
    &\Rightarrow& (\partial_{k_x}U^\dagger)\sigma_z U + U^\dagger\sigma_z(\partial_{k_x}U) = \partial_{k_x}\hat{b}\cdot\vec{\sigma} \\
    &\Rightarrow& [(\partial_{k_x}U^\dagger)U, \hat{b}\cdot\vec{\sigma}]  = \partial_{k_x}\hat{b}\cdot\vec{\sigma} \\\end{aligned}$$ In going from the first to second line, we differentiated the first line with respect to $k_x$ and then inserted $\mathbb{I}_\sigma=UU^\dagger$ in the appropriate places. From the last line, we use the fact that $[\vec{a}\cdot\vec{\sigma}, \vec{b}\cdot\vec{\sigma}] = i (\vec{a}\times\vec{b})\cdot\vec{\sigma}$ to conclude that $$-i E_x U\partial_{k_x}U^\dagger = E_x (\hat{b}\times\partial_{k_x}\hat{b})\cdot\vec{\sigma}.$$

This is the MF potential for a spin 1/2 system with an electric field in the $x$ direction.

Notice that although $U$ is not unique, the lab frame direction of $-i U\partial_{k_x}U^\dagger$ is independent of the specific choice of $U$. $E_x(-i U\partial_{k_x}U^\dagger)$ can then be thought as as an electric field dependent effective magnetization which confers a spin accumulation in the $(\hat{b}\times\partial_{k_x}\hat{b})\cdot\vec{\sigma}$ direction. Taking the specific example of the Rashba SOI where $\vec{B} = \alpha( p_y, -p_x)$ both $\vec{B}$ and $\partial_{k_x}\vec{B}$ lie on the $xy$ plane. $E_x(\hat{b}\times\partial_{k_x}\hat{b})$ thus points in the out of plane spin direction, 