Another way to see this is to use the transformation laws for the $\gamma^\mu$ matrices. Note that we can write $$\Lambda_{1/2}=\left(\begin{matrix} \Lambda_{1/2L} & 0 \\ 0 & \Lambda_{1/2R} \end{matrix}\right),$$ where $\Lambda_{1/2}$ is the matrix that performs a Lorentz transformation on a Dirac spinor. It is easy to confirm that this is true because $\Lambda_{1/2}$ is made from combinations of the identity and the block diagonal matrices $S^{\mu\nu}$ (assuming we are using the chiral representation, as in Peskin and Schroesder). It is easy to see that $\Lambda_{1/2L}$ is the operator that performs a Lorentz transformation on a left-handed spinor, and $\Lambda_{1/2R}$ does the same for a right-handed spinor. To give a concrete example, for infinitesimal transformations, equation 3.37 says that $$\Lambda_{1/2L}=1-i\theta\cdot\sigma/2-\beta\cdot\sigma/2\text{, and }\Lambda_{1/2R}=1-i\theta\cdot\sigma/2+\beta\cdot\sigma/2.$$
Now, we can use the transformation laws of the $\gamma^\mu$ matrices: $$\left(\begin{matrix} \Lambda^{-1}_{1/2L} & 0 \\ 0 & \Lambda^{-1}_{1/2R} \end{matrix}\right)\left(\begin{matrix} 0 & \sigma^\mu \\ \bar{\sigma}^\mu & 0 \end{matrix}\right)\left(\begin{matrix} \Lambda_{1/2L} & 0 \\ 0 & \Lambda_{1/2R} \end{matrix}\right)=\Lambda^{-1}_{1/2}\gamma^\mu\Lambda_{1/2}={\Lambda^\mu}_\nu\gamma^\nu=\left(\begin{matrix} 0 & {\Lambda^\mu}_\nu\sigma^\nu \\ {\Lambda^\mu}_\nu\bar\sigma^\nu & 0 \end{matrix}\right).$$$$\left(\begin{matrix} \Lambda^{-1}_{1/2L} & 0 \\ 0 & \Lambda^{-1}_{1/2R} \end{matrix}\right)\left(\begin{matrix} 0 & \sigma^\mu \\ \bar{\sigma}^\mu & 0 \end{matrix}\right)\left(\begin{matrix} \Lambda_{1/2L} & 0 \\ 0 & \Lambda_{1/2R} \end{matrix}\right)$$ $$=\Lambda^{-1}_{1/2}\gamma^\mu\Lambda_{1/2}={\Lambda^\mu}_\nu\gamma^\nu=\left(\begin{matrix} 0 & {\Lambda^\mu}_\nu\sigma^\nu \\ {\Lambda^\mu}_\nu\bar\sigma^\nu & 0 \end{matrix}\right).$$ Performing the matrix multiplication, and focusing on the top right block, we get $$\Lambda^{-1}_{1/2L}\sigma^\mu\Lambda_{1/2R}={\Lambda^\mu}_\nu\sigma^\nu.$$