Acting one basis $e^a$, The covariant derivatives \begin{align} D e^a = de^a + w^{a}{}_{b} \wedge e^b \end{align} where $w^{a}{}_{b}$ is the one-form spin connection.
I am having difficulties for expressions of $D X_{ab}$ where $X_{ab}$ $0$-form, $1$-form and $2$-form.
My idea for $X_{ab}$ on two form is \begin{align} D X^{ab} = dX^{ab} + w^{a}{}_{c} \wedge X^{cb} - X^{a}{}_{c} \wedge w^{c}{}_{b} \end{align} and acting on one-form is \begin{align} D X^{ab} = dX^{ab} + w^{a}{}_{c} \wedge X^{cb} + X^{a}{}_{c} \wedge w^{c}{}_{b} \end{align} and finally acting on zero-form is \begin{align} D X^{ab} = dX^{ab} + w^{a}{}_{c} X^{cb} - X^{a}{}_{c} w^{c}{}_{b} \end{align}
The followings are my reasoning
Recall Cartan's structure formula \begin{align} &T^a = De^a =de^a + w^a{}_{b} \wedge e^b \\ &R^{ab} = d w^{ab} + w^{a}{}_{c} \wedge w^{cb} \end{align} If we employ covariant derivatives once more, we get $D R^{ab}=0$.
\begin{align} D R^{ab} &= dR^{ab} + w^{a}{}_{c} \wedge R^{cb} - R^{a}{}_{c} \wedge w^{c}{}_{b} \\ & = (d w^{ac} \wedge w_{c}{}^{b} - w^{ac} \wedge d w_{c}{}^{b} ) + w^{a}{}_{c} \wedge ( dw^{cb} + w^{cd} \wedge w_{d}{}^{b}) - (dw^{ac} + w^{ad} \wedge w_{d}{}^{c} ) \wedge w_{cb} =0 \end{align} That's the reason why for $2$-form $X_{ab}$, I obtain
\begin{align} D X^{ab} = dX^{ab} + w^{a}{}_{c} \wedge X^{cb} - X^{a}{}_{c} \wedge w^{c}{}_{b} \end{align}
Second under the decomposition of spin-connection $w^{ab} = \mathring{w}^{ab} + \kappa^{ab}$ where $\mathring{w}$ is torsion-less part so that $de^a + \mathring{w}^{a}{}_{b} \wedge e^b=0$, and hence $T^a = \kappa^{a}{}_{b} \wedge e^b$.
In the literature, they just wrote the decomposition $R^{ab} = \mathring{R}^{ab} + \mathring{D} \kappa^{ab} + \kappa^{a}{}_{c} \wedge \kappa^{c}{}_{b}$. Here $\kappa_{ab}$ is one-form. Hence the direct computations gives \begin{align} D^2 e^a &= R^{a}{}_{b} \wedge e^b = (d w^{a}{}_{b} + w^{a}{}_{c} \wedge w^{c}{}_{b} ) \wedge e^b \\ &= \left( \mathring{R}^{a}{}_{b} + d \kappa^{a}{}_{b} + \mathring{w}^{a}{}_{c} \wedge \kappa^{c}{}_{b} + \kappa^{a}{}_{c} \wedge \mathring{w}^{c}{}_{b} + \kappa^{a}{}_{c} \wedge \kappa^{c}{}_{b} \right) \wedge e^b \\ & = \left( \mathring{R}^{a}{}_{b} + \mathring{D} \kappa^{a}{}_{b} + \kappa^{a}{}_{c} \wedge \kappa^{c}{}_{b} \right) \wedge e^b \end{align} So I guess for one-form $X_{ab}$ \begin{align} D X^{ab} = dX^{ab} + w^{a}{}_{c} \wedge X^{cb} + X^{a}{}_{c} \wedge w^{c}{}_{b} \end{align}
Is my approach correct? then How one can extend this to arbitrary forms with an arbitrary index?
Attempt to make things compatible which one can read in various places in the internet such as this and Wikipedia.
The spin connection ${\boldsymbol{\omega}^a}_b={{\boldsymbol{\omega}_\mu}^a}_b\,dx^\mu$ is a $\mathfrak{so}(3,1)$-Lie-algebra valued one form. That is: for each general coordinate index $\mu$ (Greek letter), ${{\boldsymbol{\omega}_\mu}^a}_b$ is a certain $(4\times 4)$-matrix in its Lorentz indices $a,b$ (Latin letters). Alternatively, one can view ${\boldsymbol{\omega}^a}_b$ as a $(4\times 4)$-matrix of ordinary one-forms.
The indices of ${\boldsymbol{\omega}^a}_b$ are raised and lowered with the Lorentz metric $\eta={\rm diag}(1,1,1,-1)\,,$ $\boldsymbol{\omega}_{ab}=\eta_{ac}\,{\boldsymbol{\omega}^c}_b$, and not with the possibly curved metric $g_{\mu\nu}\,.$ Then there is anti-symmetry $\boldsymbol{\omega}_{ab}=-\boldsymbol{\omega}_{ba}\,.$
The coordinate basis one forms $dx^\mu$ are related to basis one forms $\boldsymbol{e}^a$ in the local Lorentz frame by $\boldsymbol{e}^a={\boldsymbol{e}^a}_\mu\,dx^\mu\,.$ A vector field $\boldsymbol{X}$ can therefore have Lorentz components $X^a=\boldsymbol{e}^a(\boldsymbol{X})$ or ordinary coordinate components $X^\mu=dx^\mu(\boldsymbol{X})\,.$ This vector field can be viewed as a vector-valued $0$-form. Its covariant derivative coming from the spin connection is therefore the Lorentz-vector valued one-form $$ DX^a=dX^a+{\boldsymbol{\omega}^a}_b\,X^b\,. $$ Applying this to the coordinate basis vector $\boldsymbol{e}_\mu$ gives $$ D_\mu X^a=\partial_\mu X^a+{{\boldsymbol{\omega}_\mu}^a}_b\,X^b\,. $$ Analogously, when $T^{ab}$ is a contravariant rank-$2$ tensor then \begin{align} D\, T^{ab}&=dT^{ab}+{\boldsymbol{\omega}^a}_c\,T^{cb}+ {\boldsymbol{\omega}^b}_c\,T^{ac}\,,\\[2mm] D_\mu\, T^{ab}&=\partial_\mu T^{ab}+{{\boldsymbol{\omega}_\mu}^a}_c\,T^{cb}+ {{\boldsymbol{\omega}_\mu}^b}_c\,T^{ac}\,. \end{align} The formulas are fully analogous to the Levi-Civita connection where the ordinary Christoffel symbols are replaced by ${{\boldsymbol{\omega}_\mu}^a}_b\,.$
Now we go from $0$-forms to one-forms. If $\boldsymbol{\alpha}^a={\boldsymbol{\alpha}^a}_b\,\boldsymbol{e}^b$ is a Lorentz-vector valued one-form then the covariant derivative is defined as $$ D\,\boldsymbol{\alpha}^a=d\boldsymbol{\alpha}^a+{\boldsymbol{\omega}^a}_b\color{red}{\wedge} \boldsymbol{\alpha}^b\,. $$ Same rule, just that the product in $\mathbb R$ is replaced by the $\color{red}{wedge~product}$. When $\boldsymbol{\alpha}^{ab}$ is a rank-$2$ Lorentz-tensor-valued one-form then (fully compatible with the case for $T^{ab}$) $$ D\,\boldsymbol{\alpha}^{ab}=d\boldsymbol{\alpha}^{ab}+{\boldsymbol{\omega}^a}_c\wedge\boldsymbol{\alpha}^{cb}\color{red}{+} {\boldsymbol{\omega}^b}_c\wedge\boldsymbol{\alpha}^{ac}\,. $$ When $\boldsymbol{\alpha}$ has one contravariant upper and one covariant lower index the $\color{red}{+}$ sign becomes a minus sign, or which is the same, we can swap the factors in the wedge product and keep the plus sign: $$ D\,{\boldsymbol{\alpha}^a}_b=d{\boldsymbol{\alpha}^a}_b+{\boldsymbol{\omega}^a}_c\wedge{\boldsymbol{\alpha}^c}_b+ {\boldsymbol{\alpha}^a}_c\wedge {\boldsymbol{\omega}^c}_b\,. $$
All of the above sign conventions are compatible with the formula in Wikipedia by which the covariant derivative of a $(1,1)$-tensor valued $p$-form is $$\boxed{\quad\phantom{\Bigg|} D{V^a}_b=d{V^a}_b+{\boldsymbol{\omega}^a}_c\wedge {V^c}_b-(-1)^p\,{V^a}_c\wedge {\boldsymbol{\omega}^c}_b\,.\quad} $$ When $V$ has two upper indices (like your $X$) we should have $$\boxed{\quad\phantom{\Bigg|} DV^{ab}=dV^{ab}+{\boldsymbol{\omega}^a}_c\wedge V^{cb}\color{red}{+}(-1)^p\,V^{ac}\wedge {\boldsymbol{\omega}^b}_c\,.\quad} $$ Swapping the wedge products allows to get rid of the $(-1)^p\,:$ $$\boxed{\quad\phantom{\Bigg|} D{V^a}_b=d{V^a}_b+{\boldsymbol{\omega}^a}_c\wedge {V^c}_b- {\boldsymbol{\omega}^c}_b\wedge {V^a}_c\,.\quad} $$ $$\boxed{\quad\phantom{\Bigg|} DV^{ab}=dV^{ab}+{\boldsymbol{\omega}^a}_c\wedge V^{cb}\color{red}{+} {\boldsymbol{\omega}^b}_c\wedge\,V^{ac}\,.\quad} $$ In that format the signs only depend on the index placements and not what rank $p$ that form $V$ has.