= Solution
Use signature $\eta=\operatorname{diag}(1,-1,-1,-1)$, the numerical <matrices> $\sigma_\mu=(I,\sigma_i)$ and $\bar\sigma_\mu=(I,-\sigma_i)$ given in the question, and raise genuine <tensor> indices with $\eta$. A real <Minkowski spacetime> vector corresponds to the <Hermitian matrix>
$$
X=x^0I+\mathbf x\cdot\boldsymbol\sigma,\qquad \det X=(x^0)^2-|\mathbf x|^2.
$$
For <SL(2,C)> <matrices>, $X'=AXA^\dagger$ is Hermitian and has the same <determinant>. It therefore induces a real linear <Lorentz transformation>; composition of the congruence actions agrees with <matrix> multiplication. The kernel consists of $\pm I$: preservation of $X=I$ makes a kernel element unitary, and preservation of every Hermitian $X$ makes it scalar.
<Polar decomposition of an invertible complex matrix> connects every determinant-one complex <matrix> to its unitary factor, so the group is connected. The action gives the <Lorentz spinor double cover> of the <proper orthochronous Lorentz group>. It covers rotations via $A_R=e^{-i\vartheta\mathbf n\cdot\boldsymbol\sigma/2}$ and boosts via the positive Hermitian <matrices> below. Every proper orthochronous transformation is a boost followed by a rotation, since one can first match its image of the future unit time vector and then use its rotation stabilizer. \b[The congruence construction does not cover spatial parity or time reversal.] In particular parity has <determinant> $-1$ as a four-vector transformation; all transformations continuously produced from <SL(2,C)> have <determinant> $+1$.
Set $c_\theta=\cosh(\theta/2)$, $s_\theta=\sinh(\theta/2)$ and $N=\mathbf n\cdot\boldsymbol\sigma$. The <Pauli matrix multiplication law> gives $N^2=I$. The proposed boost <matrix> is $A_B=c_\theta I+s_\theta N$, with <eigenvalues> $e^{\pm\theta/2}$, <determinant> one and $A_B^\dagger=A_B$. Split $\mathbf x=x_\parallel\mathbf n+\mathbf x_\perp$. The perpendicular Pauli part anticommutes with $N$, so multiplication gives
$$
\boxed{x'^0=\cosh\theta\,x^0+\sinh\theta\,x_\parallel,\qquad
x'_\parallel=\sinh\theta\,x^0+\cosh\theta\,x_\parallel,\qquad
\mathbf x'_\perp=\mathbf x_\perp.}
$$
This is an active <Lorentz boost>. The image of a rest worldline has velocity \b[$\mathbf v=\tanh\theta\,\mathbf n$], fixing the velocity sign convention.
For two nonzero boosts, write $A'=c'I+s'\mathbf n'\cdot\boldsymbol\sigma$ and $A=cI+s\mathbf n\cdot\boldsymbol\sigma$. Their product is
$$
A'A=(c'c+s's\mathbf n'\cdot\mathbf n)I
+(c's\mathbf n+s'c\mathbf n')\cdot\boldsymbol\sigma
+i s's(\mathbf n'\times\mathbf n)\cdot\boldsymbol\sigma.
$$
The last term is anti-Hermitian. Thus $A'A$ is not Hermitian unless the axes are collinear; a pure boost has Hermitian lifts $\pm A_B$, so it cannot give the same Lorentz transformation. The <noncollinear boost obstruction from Pauli products> is therefore
$$
\boxed{\text{two nonzero noncollinear boosts require an accompanying rotation}.}
$$
That rotation is a <Wigner rotation>. A zero-rapidity factor is the trivial exception, regardless of the arbitrary axis assigned to it.
Define $J_i=\tfrac12\epsilon_{ijk}M_{jk}$ and $K_i=M_{0i}$. The printed <Lorentz algebra> brackets give
$$
[J_i,J_j]=i\epsilon_{ijk}J_k,\qquad
[J_i,K_j]=i\epsilon_{ijk}K_k,\qquad
[K_i,K_j]=-i\epsilon_{ijk}J_k.
$$
Set $L_i=(J_i+iK_i)/2$, $R_i=(J_i-iK_i)/2$. Then
$$
\boxed{[L_i,L_j]=i\epsilon_{ijk}L_k,\quad
[R_i,R_j]=i\epsilon_{ijk}R_k,\quad [L_i,R_j]=0.}
$$
This is the <chiral decomposition of the complex Lorentz algebra>. The two copies are the complexifications of the <SU(2)> algebras conventionally labelled left and right. They are not two independent compact real subalgebras of the real Lorentz algebra: $L,R$ are complex linear combinations of the real generators. The distinction is needed for noncompact boosts.
To verify the infinitesimal two-component action, put
$$
a=\frac14\omega^{\mu\nu}\sigma_\mu\bar\sigma_\nu,\qquad A=I+a+O(\omega^2).
$$
Its trace vanishes by antisymmetry, so $\det A=1+O(\omega^2)$. The <Pauli matrix multiplication law> implies
$$
\sigma_\mu\bar\sigma_\nu\sigma_\rho+\sigma_\rho\bar\sigma_\nu\sigma_\mu
=2(\eta_{\mu\nu}\sigma_\rho+\eta_{\nu\rho}\sigma_\mu-\eta_{\mu\rho}\sigma_\nu).
$$
Since $a^\dagger=\tfrac14\omega^{\mu\nu}\bar\sigma_\nu\sigma_\mu$, antisymmetrizing this identity yields
$$
a\sigma_\rho+\sigma_\rho a^\dagger
=\omega^\mu{}_{\rho}\sigma_\mu.
$$
Consequently $AXA^\dagger=X+\sigma_\mu\omega^\mu{}_{\nu}x^\nu+O(\omega^2)$, exactly the claimed infinitesimal coordinate transformation with $\omega^\mu{}_{\nu}=\omega^{\mu\rho}\eta_{\rho\nu}$.
The two <Weyl spinor> representations transform as
$$
\boxed{\psi_L\mapsto A\psi_L,\qquad \psi_R\mapsto(A^\dagger)^{-1}\psi_R.}
$$
Both maps preserve group multiplication. For $\Lambda=e^\omega$, choose its spin lift and write the finite <matrices> as
$$
A=\exp\left(\frac14\omega^{\mu\nu}\sigma_\mu\bar\sigma_\nu\right),\qquad
(A^\dagger)^{-1}=\exp\left(\frac14\omega^{\mu\nu}\bar\sigma_\mu\sigma_\nu\right).
$$
For rotations these coincide as the <SU(2)> doublet; for boosts their generator signs are opposite. Any complex-linear intertwiner would commute with all rotations and hence be scalar by the <Schur lemma>, but a nonzero scalar cannot intertwine the opposite boost <matrices>. They are therefore inequivalent. Infinitesimally their $K_i$ <matrices> are $-i\sigma_i/2$ and $+i\sigma_i/2$, while their $J_i$ <matrices> are both $\sigma_i/2$. Thus they have chiral labels $(1/2,0)$ and $(0,1/2)$. The finite <matrices> are representations of the spin cover; choosing $A$ or $-A$ matters for spinors even though their four-vector transformations coincide.
Using the supplied block <gamma matrices>, the generator of the <Dirac spinor> is
$$
\Omega=\frac14\omega^{\mu\nu}\gamma_{\mu\nu}
=\begin{pmatrix}\tfrac14\omega^{\mu\nu}\sigma_\mu\bar\sigma_\nu&0\\0&\tfrac14\omega^{\mu\nu}\bar\sigma_\mu\sigma_\nu\end{pmatrix}.
$$
The antisymmetry of $\omega$ removes the symmetric Clifford part. Hence
$$
\boxed{\psi\mapsto S\psi,\qquad S=e^\Omega=\operatorname{diag}(A,(A^\dagger)^{-1}).}
$$
There is a conjugation-order error in the last displayed gamma identity in the PDF. The <Clifford algebra> gives, with $\gamma^\rho=\eta^{\rho\sigma}\gamma_\sigma$,
$$
[\gamma_{\mu\nu},\gamma^\rho]=2(\delta_\nu^\rho\gamma_\mu-\delta_\mu^\rho\gamma_\nu),\qquad
[\Omega,\gamma^\rho]=-\omega^\rho{}_{\mu}\gamma^\mu.
$$
Exponentiating this linear <commutator> action proves the <inverse Lorentz action on gamma matrices>
$$
\boxed{S\gamma^\rho S^{-1}=(\Lambda^{-1})^\rho{}_{\mu}\gamma^\mu,\qquad
S^{-1}\gamma^\rho S=\Lambda^\rho{}_{\mu}\gamma^\mu.}
$$
The second identity is the order required for the requested bilinears when $\psi\mapsto S\psi$. An explicit countercheck to the printed order is a positive boost along the third axis: $\omega^{03}=-\theta$ gives
$$
S\gamma^0S^{-1}=\cosh\theta\,\gamma^0-\sinh\theta\,\gamma^3,
$$
whereas the printed right side has a plus sign. This cannot be repaired by dropping index raising; the same raising convention is needed in the preceding coordinate transformation.
The adjoint relation supplied in the question gives $\Omega^\dagger\gamma^0=-\gamma^0\Omega$, hence the <Dirac spinor pseudo-unitarity> identity $S^\dagger\gamma^0S=\gamma^0$. The <Dirac adjoint> therefore transforms as $\bar\psi\mapsto\bar\psi S^{-1}$. It now follows that the <Dirac scalar bilinear> is
$$
\boxed{\bar\psi'\psi'=\bar\psi S^{-1}S\psi=\bar\psi\psi,}
$$
the vector is
$$
\boxed{\bar\psi'\gamma^\rho\psi'=\Lambda^\rho{}_{\mu}\bar\psi\gamma^\mu\psi,}
$$
and the antisymmetric second-rank <tensor> is
$$
\boxed{\bar\psi' i[\gamma^\rho,\gamma^\sigma]\psi'
=\Lambda^\rho{}_{\mu}\Lambda^\sigma{}_{\nu}\bar\psi i[\gamma^\mu,\gamma^\nu]\psi.}
$$
These are respectively a <Lorentz scalar>, <Lorentz four-vector> and <Lorentz tensor>. The source's inconsistent gamma identity is replaced by its correct inverse/order pair; all three transformation laws then follow with the stated spinor transformation.
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