Ricci identity:
$\forall X, Y \in \Gamma(TM), T \in\Gamma(\otimes^{r,s}TM)$, we have
$$\nabla^2T(\cdots,X,Y)-\nabla^2T(\cdots,Y,X)=-R(X,Y)T(\cdots)$$
where $R$ is curvature tensor, $R(X,Y):=\nabla_X \nabla_Y - \nabla_Y \nabla_X-\nabla_{[X,Y]}$, $\nabla$ is torsion free connection.
I want to check Ricci Identity for $T \in\Gamma(\otimes^{1,0}TM)$, i.e.
$$\nabla^2T(\omega,X,Y)-\nabla^2T(\omega,Y,X)=-R(X,Y)T(\omega)$$
My effort:
$\forall X,Y,\in \Gamma(TM), \omega \in \Gamma(T^*M)$, we have
\begin{align} \nabla^2T(\omega,X,Y)&=\nabla(\nabla T)(\omega,X,Y)=(\nabla_Y(\nabla T))(\omega,X)\\ &=\nabla_Y(\nabla T(\omega,X))-(\nabla T)(\nabla_Y \omega,X)-(\nabla T)(\omega,\nabla_Y X)\\ &=\nabla_Y \nabla_X T(\omega)-\nabla_X T(\nabla_Y \omega)-(\nabla_{\nabla_Y X} T)(\omega)\text{ (**wrong**)} \end{align}
If we want $\nabla^2T(\omega,X,Y)-\nabla^2T(\omega,Y,X)=-R(X,Y)T(\omega)$
$\quad$ we need $\nabla_X T(\nabla_Y \omega)=\nabla_Y (\nabla_X \omega)$. Are these two things equal?
Update:
As mentioned in answer by Arctic Char, there's a mistake in calculation. The correct one should be:
\begin{align} \nabla^2T(\omega,X,Y)&=\nabla(\nabla T)(\omega,X,Y)=(\nabla_Y(\nabla T))(\omega,X)\\ &=\nabla_Y(\nabla T(\omega,X))-(\nabla T)(\nabla_Y \omega,X)-(\nabla T)(\omega,\nabla_Y X)\\ &=\nabla_Y (\nabla_X T(\omega))-\nabla_X T(\nabla_Y \omega)-(\nabla_{\nabla_Y X} T)(\omega)\\ &=\nabla_Y \nabla_X T(\omega)+\nabla_X T(\nabla_Y \omega)-\nabla_X T(\nabla_Y \omega)-(\nabla_{\nabla_Y X} T)(\omega)\\ &=\nabla_Y \nabla_X T(\omega)-(\nabla_{\nabla_Y X} T)(\omega) \end{align}
