= Solution
Use coordinate indices $\mu,\nu$ and Lorentz-frame indices $A,B$, and take $\eta_{AB}=\operatorname{diag}(-1,1,1,1)$. Locally choose four independent one-forms $e^A=e^A{}_\mu dx^\mu$ satisfying
$$
\boxed{g_{\mu\nu}=\eta_{AB}e^A{}_\mu e^B{}_\nu.}
$$
One construction begins with a smooth timelike vector field in a sufficiently small neighborhood, normalizes it, and applies the <Gram-Schmidt process> to its spacelike orthogonal complement. The dual <orthonormal coframe in spacetime> gives the <vierbein>; equivalently its matrix factorizes the metric as $g=e^T\eta e$. The inverse matrix gives the frame $e_A=e_A{}^\mu\partial_\mu$ with $e_A{}^\mu e^B{}_\mu=\delta_A^B$. This is a local construction: a global frame is not guaranteed by a metric, and a globally defined <spinor field> requires a compatible <spin structure>.
The metric does not determine a unique <vierbein>. A <local Lorentz transformation> changes the coframe components by
$$
\boxed{e'^A{}_\mu=\Lambda^A{}_B(x)e^B{}_\mu,\qquad
\Lambda^T\eta\Lambda=\eta,\qquad
e'_A{}^\mu=(\Lambda^{-1})^B{}_A e_B{}^\mu.}
$$
The coordinate index is unchanged in this frame transformation. Restrict to the proper orthochronous Lorentz group for its connected spin lift; orientation-reversing transformations require further structure and are not part of this local spin-group discussion.
Let constant <gamma matrices> obey the <Clifford algebra> $\{\gamma^A,\gamma^B\}=2\eta^{AB}$, and write $\gamma^{AB}=\tfrac12[\gamma^A,\gamma^B]$. A <spinor field> transforms by the spin lift of the Lorentz transformation, not by its vector matrix:
$$
\boxed{\psi'(x)=S(\Lambda(x))\psi(x),\qquad
S^{-1}\gamma^A S=\Lambda^A{}_B\gamma^B.}
$$
For $\Lambda=1+\alpha+O(\alpha^2)$, with $\alpha_{AB}=-\alpha_{BA}$, the <Clifford algebra> identity
$$
[\gamma^C,\gamma^{AB}]=2(\eta^{CA}\gamma^B-\eta^{CB}\gamma^A)
$$
shows explicitly that
$$
S=1+\frac14\alpha_{AB}\gamma^{AB}+O(\alpha^2).
$$
Indeed the commutator of $\gamma^C$ with this infinitesimal spin matrix is $\alpha^C{}_B\gamma^B$. A finite transformation connected to the identity is obtained by exponentiating these generators along a path. Its lift has two possible signs, so spinors are not ordinary Lorentz vectors: a $2\pi$ spatial rotation acts by $-1$, and $4\pi$ by $+1$. Consistent transition lifts define the <spin structure>; the metric alone does not select it.
For the metric connection, take the <Levi-Civita connection> and determine the <spin connection> from the tetrad postulate
$$
\partial_\mu e^A{}_\nu+\omega_\mu{}^A{}_B e^B{}_\nu-\Gamma^\rho_{\mu\nu}e^A{}_\rho=0.
$$
It is equivalently determined by $de^A+\omega^A{}_B\wedge e^B=0$ and $\omega_{\mu AB}=-\omega_{\mu BA}$. Explicitly, in terms of the inverse frame,
$$
\omega_\mu{}^A{}_B=e^A{}_\nu\left(\partial_\mu e_B{}^\nu+\Gamma^\nu_{\mu\rho}e_B{}^\rho\right).
$$
Thus both the coordinate connection and the Lorentz <connection 1-form> are determined by the metric and chosen frame; no independent torsion has been supplied.
An ordinary derivative fails covariance because $\partial_\mu(S\psi)=S\partial_\mu\psi+(\partial_\mu S)\psi$. Seek $\nabla_\mu\psi=(\partial_\mu+\Omega_\mu)\psi$. Covariance requires
$$
\Omega'_\mu=S\Omega_\mu S^{-1}-(\partial_\mu S)S^{-1}.
$$
The coframe's connection obeys $\omega'=\Lambda\omega\Lambda^{-1}-d\Lambda\,\Lambda^{-1}$. Applying the spin representation of the same Lorentz algebra, whose infinitesimal generator was just determined, supplies
$$
\boxed{\nabla_\mu\psi=\partial_\mu\psi+\frac14\omega_{\mu AB}\gamma^{AB}\psi
=\partial_\mu\psi+\frac18\omega_{\mu AB}[\gamma^A,\gamma^B]\psi.}
$$
This proves the transformation law rather than choosing the coefficient arbitrarily. It is the <spinor covariant derivative> induced by the <spin connection>. With curved matrices $\gamma^\nu(x)=e_A{}^\nu\gamma^A$, the same coefficient and tetrad postulate give
$$
\partial_\mu\gamma^\nu+\Gamma^\nu_{\mu\rho}\gamma^\rho+[\Omega_\mu,\gamma^\nu]=0,
$$
so differentiation is compatible with Clifford multiplication and $\{\gamma^\mu(x),\gamma^\nu(x)\}=2g^{\mu\nu}(x)$.
Finally, expand the commutator of these derivatives:
$$
[\partial_\mu+\Omega_\mu,\partial_\nu+\Omega_\nu]\psi
=(\partial_\mu\Omega_\nu-\partial_\nu\Omega_\mu+[\Omega_\mu,\Omega_\nu])\psi.
$$
The spin representation preserves the Lorentz algebra bracket. Consequently $d\Omega+\Omega\wedge\Omega=\tfrac14R_{AB}\gamma^{AB}$, where $R^A{}_B=d\omega^A{}_B+\omega^A{}_C\wedge\omega^C{}_B$. In components the <spinor curvature identity> is
$$
\boxed{[\nabla_\mu,\nabla_\nu]\psi=\frac14R_{\mu\nu AB}\gamma^{AB}\psi.}
$$
For the full second derivative, $\nabla_\mu(\nabla_\nu\psi)$ also includes $-\Gamma^\rho_{\mu\nu}\nabla_\rho\psi$, because its first derivative has a covector index. This term cancels on antisymmetrization for the <torsion-free connection>. With unit-weight brackets, $X_{[\mu\nu]}=(X_{\mu\nu}-X_{\nu\mu})/2$, the requested expression is therefore
$$
\boxed{\nabla_{[\mu}\nabla_{\nu]}\psi
=\frac18R_{\mu\nu AB}\gamma^{AB}\psi
=\frac1{16}R_{\mu\nu AB}[\gamma^A,\gamma^B]\psi.}
$$
This convention distinguishes the antisymmetrized second derivative from the full commutator by its factor of two. In a noncoordinate orthonormal frame, an operator commutator must additionally subtract the derivative along the frame-vector commutator; using the full tensorial second derivative incorporates that correction.
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