= Solution
Insert $\Omega_{jk}=-\epsilon_{jkl}\phi^l$ and the result for $S^{jk}$ into the <Spinor representation of the Lorentz group>. The <contraction of two Levi-Civita symbols> $\sum_{j,k}\epsilon_{jkl}\epsilon^{jkm}=2\delta_l{}^m$ gives
$$
\frac12\Omega_{jk}S^{jk}=\frac i2\begin{pmatrix}\boldsymbol\phi\cdot\boldsymbol\sigma&0\\0&\boldsymbol\phi\cdot\boldsymbol\sigma\end{pmatrix}.
$$
The <matrix exponential> of this block diagonal matrix is the stated $O$, so $\psi'(x)=O\psi(\Lambda^{-1}x)$. The plus sign in its exponent follows from both minus signs above; it is tied to the rotation-parameter convention in the PDF.
For a rotation through $2\pi$ about $z$, the <Pauli matrix> has <eigenvalues> $\pm1$, hence $e^{i\pi\sigma^3}=-I_2$. The spacetime rotation is the identity, while
$$
\boxed{\psi'(x)=-\psi(x)\quad\text{for a }2\pi\text{ rotation};\qquad\psi'(x)=\psi(x)\quad\text{for }4\pi.}
$$
This is the <spinor sign under a full spatial rotation>. A <vector> returns to itself already at $2\pi$. The comparison concerns the exponentiated <Lorentz group> representations: a <Lie algebra> by itself has no distinguished element called a $2\pi$ rotation. The sign records the nontrivial lift of this rotation loop to the <Spin group>.
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