Solutions of the Dirac equation
To obtain solutions for the spinors, we solve the Dirac equation $$(i\cancel{\partial}-m)\psi=0.$$ By multiplying with $(i\cancel{\partial}+m)$ we obtain $$(\Box+m^2)\psi=0,$$ which means the spinors also satisfy the Klein-Gordon equation.
The solutions to the Klein-Gordon equation can be written in terms of plane waves $$\psi_s(x)=\int\frac{d^3p}{(2\pi)^3}u_s(p)e^{-ipx},\qquad \chi_s(x)=\int\frac{d^3p}{(2\pi)^3}v_s(p)e^{ipx}$$ The $u_s(p)$ and $v_s(p)$ are the polarizations for patricles and antiparticles. We need to finde them for fixed $p^\mu$ like in the Spin $1$ case, by inserting into the dirac equation itself. They then transform under the Poincare group through the transformation of $p^\mu$.
Inserting the Spinors into the dirac Dirac equation gives $$\begin{pmatrix}-m & p\cdot\sigma\\ p\cdot\bar{\sigma} & -m\end{pmatrix}u_s(p)=\begin{pmatrix}-m& -p\cdot\sigma \\ -p\cdot\bar{\sigma} & -m\end{pmatrix}v_s(p)=0$$ We can choose a fixed $p^\mu$, e.g. $p^\mu = (m,0,0,0)$, which leads to $$\begin{pmatrix}-1 & 1\\ 1 & -1\end{pmatrix}u_s(p)=\begin{pmatrix}-1& -1 \\ -1 & -1\end{pmatrix}v_s(p)=0$$ This leads to the solutions $$u_s=\begin{pmatrix}\xi_s\\xi_s\end{pmatrix},\quad v_s=\begin{pmatrix}\eta_s\\ -\eta_s\end{pmatrix},$$ with arbitrary two component spinors $\xi_s$ and $\eta_s$.
Since $s$ are the spin degrees of freedom $s=\uparrow,\downarrow$, we have four independent basis vectors $$u_\uparrow = \sqrt{m}\begin{pmatrix}1\0\1\0\end{pmatrix},\quad u_\downarrow=\sqrt{m}\begin{pmatrix}0\1\0\1\end{pmatrix},\quad v_\uparrow=\sqrt{m}\begin{pmatrix}-1\0\1\0\end{pmatrix},\quad v_\downarrow=\sqrt{m}\begin{pmatrix}0\1\0\-1\end{pmatrix}$$ Initially the Dirac spinor had eight real degrees of freedom (four complex) but the dirac equation reduced it to four thus four independent polarization vectors.
In a different frame, where $p^\mu=(E,0,0,p_z)$ we have $$u_s(p)=\begin{pmatrix}\sqrt{p\cdot\sigma}\xi_s\\ \sqrt{p\cdot\bar{\sigma}}\xi_s\end{pmatrix},\quad u_s(p)=\begin{pmatrix}\sqrt{p\cdot\sigma}\eta_s\\ -\sqrt{p\cdot\bar{\sigma}}\eta_s\end{pmatrix},$$ with $$\sqrt{p\cdot\sigma}=\begin{pmatrix}\sqrt{E-p_z}&0\0&\sqrt{E+p_z}\end{pmatrix},\quad \sqrt{p\cdot\bar{\sigma}}=\begin{pmatrix}\sqrt{E+p_z} & 0 \\ 0 & \sqrt{E-p_z}\end{pmatrix}.$$
Choosing $\xi_s$ and $\eta_s$ we get four solutions $$u^1_p=\begin{pmatrix}\sqrt{E-p_z}\\ 0\\ \sqrt{E+p_z}\0\end{pmatrix},;u^2_p=\begin{pmatrix}0\\sqrt{E+p_z}\\ 0\\ \sqrt{E-p_z}\end{pmatrix},;v^1_p=\begin{pmatrix}\sqrt{E-p_z}\\ 0\\ -\sqrt{E+p_z}\0\end{pmatrix},;v^2_p=\begin{pmatrix}0\\sqrt{E+p_z}\\ 0\\ -\sqrt{E-p_z}\end{pmatrix}.$$
In any frame the $u$s correspond to electrons (because they have positive frequency) and the $v$s correspond to positrons (because they have negative frequency).
The full dirac spinor $\psi$ is a sum over the different polarizations, therefore $\psi=(\psi_L,\psi_R)$ gets contributions from both particles and anti-particles.
One finds that the above definition leads to the normalization $$\bar{u}_s(p)u_{s’}(p) = u^\dagger_s(p)\gamma_0u_{s’}(p)=2m\delta_{ss’}$$ $$\bar{v}_s(p)v_{s’}(p)=-2m\delta_{ss’}$$ $$\bar{v}_s(p)u_{s’}(p)=\bar{u}_s(p)v_{s’}(p)=0$$
The outer product of the spinors gives the completeness relations
Spinor completeness relations
$$\sum^2_{s=1}u_s(p)\bar{u}_s(p)=\cancel{p}+m$$ $$\sum^2_{s=1}v_s(p)\bar{v}_s(p)=\cancel{p}-m$$