I will be using a different notation to you, and all of what I write down may be found in Ref. [1] chapters 1.1 - 1.6, I highly recommend it (section 1.6 is particularly relevant, but you will need the prev sections too).
I will do everything in the tangent frame (so all indices are local Lorentz), you can just convert back using the vielbein if you want. I will also be working in four dimensions. The covariant derivative is written as
$$\nabla_{a}=e_a+\frac{1}{2}\omega_a{}^{bc}M_{bc}~,$$
where $e_a=e_a{}^{m}\partial_m$ is the inverse vielbein, $\omega_a{}^{bc}$ is the spin connection and $M_{ab}$ are the Lorentz generators. The important difference between my expression and yours is that my Lorentz generators are in an arbitrary representation, where as yours are in what I think might be the Dirac representation (i.e. 4 component reducible spinors).
One can then show that, in the torsion free case, the commutator of covariant derivatives is
$$[\nabla_a,\nabla_b]=\frac{1}{2}R_{ab}{}^{cd}M_{cd}~.$$
The beauty of this formula is that it can act on any type of spin-tensor, you only need to remember how the Lorentz generators act on different objects, the Leibniz rule for the generators will take you from there.
I will give some examples on how $M_{ab}$ act on certain objects, but you are probably familiar with these. Let $V_a$ be a vector, $\psi_{\alpha}$ and $\bar{\chi}^{\dot{\alpha}}$ a left and right handed 2 component spinor respectively and let $\Psi=\big(\psi_{\alpha},~\bar{\chi}^{\dot{\alpha}}\big)^T$ be a 4 component spinor. The Lorentz generators act on the as follows
\begin{align}
M_{ab}V_c&=\eta_{ca}V_b-\eta_{cb}V_a~, \tag{1.a}\\
M_{ab}\psi_{\alpha}&=(\sigma_{ab})_{\alpha}{}^{\beta}\psi_{\beta}~, \tag{1.b}\\
M_{ab}\bar{\chi}^{\dot{\alpha}}&=(\tilde{\sigma}_{ab})^{\dot{\alpha}}{}_{\dot{\beta}}\bar{\chi}^{\dot{\beta}}~,\tag{1.c}\\
M_{ab}\Psi&=\Sigma_{ab}\Psi~. \tag{1.d}
\end{align}
Here $\sigma_{ab}=-\frac{1}{4}\big(\sigma^a\tilde{\sigma}^b-\sigma^b\tilde{\sigma}^a\big)$, $\Sigma_{ab}=-\frac{1}{4}[\gamma_a,\gamma_b]$ and $\gamma_a=\begin{pmatrix} 0 & \sigma_a \\ \tilde{\sigma}_a &0\end{pmatrix}$ etc etc.
Of course, if you are working with spinors, it is usually much easier to convert everything into 2 component spinor notation, then there is only two rules to remember.
To conclude I will do one of your examples explicitly. I will assume that your $\psi$ is a 4 component spinor.
\begin{align}
[\nabla_a,\nabla_b]\nabla_c\nabla_d\psi&=\frac{1}{2}R_{ab}{}^{fg}M_{fg}\bigg(\nabla_c\nabla_d\psi\bigg) \\
&=\frac{1}{2}R_{ab}{}^{fg}\bigg(M_{fg}\nabla_c\nabla_d\psi+\nabla_cM_{fg}\nabla_d\psi+\nabla_c\nabla_dM_{fg}\psi\bigg)\\
&=\frac{1}{2}R_{ab}{}^{fg}\bigg(2\eta_{cf}\nabla_g\nabla_d\psi+2\nabla_c\eta_{df}\nabla_g\psi+\nabla_c\nabla_d\Sigma_{fg}\psi\bigg)\\
&=R_{abc}{}^{f}\nabla_f\nabla_d\psi+R_{abd}{}^{f}\nabla_c\nabla_f\psi+\frac{1}{2}R_{ab}{}^{fg}\Sigma_{fg}\nabla_c\nabla_d\psi~.
\end{align}
[1] I.L. Buchbinder and S.M. Kuzenko, Ideas and Methods of Supersymmetry and Supergravity, Or a Walk Through Superspace, IOP, Bristol (1995) (Revised Edition 1998).