Spinor form of the exterior derivative

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In the book of Ward and Wells "Twistor Geometry and Field Theory" and also in the paper "Cohomology and Massless Fields" of Eastwood, Penrose and Wells, appears a spinor form of the exterior derivative $d$. This is the operator $\nabla_{AA'}:\mathcal O\rightarrow\mathcal O_{AA'}$. I think that this is just an extension of the opertor $d$ to act in spinors and tensor fields, but I don't know how to give a sense to this, i.e., I don't know how this derivative acts in spinors that can't be expressed as tensors.

Also, identifying the 1-forms in the Grassmannian $G_2(\mathbb C^4)$, $\Omega^1$ with $\mathcal O_{AA'}$, it is said that the exterior derivative splits $\Omega^2$ in its self-dual and anti-self-dual part

$$ \psi_{AA'}\rightarrow\nabla_{\left(B'\right.}^A\psi_{\left.A'\right)A}+\nabla_{\left(B\right.}^{A'}\psi_{\left.A\right)A'}. $$ Where $\nabla_{A}^{B'}=\nabla_{AA'}\varepsilon^{A'B'}$ and $\nabla_{B'}^{A}=\varepsilon^{AB}\nabla_{BB'}$ and $\varepsilon_{AB}$ is the sympletic form acting on the spin space.

I understand that $d$ splits the two-forms in the self-dual and anti-self-dual parts, but I don't see how the derivative in the sense of spinor is giving me that last expression. My problem is that I don't understand this derivative $\nabla_{AA'}$, Could you help to "see" how this differential operator is acting on spinors?