(I'm new here and I hope it's fine to ask a GR/diff geometry question since that is a grad level discussion)

I'm reading Sean Carroll's notes on GR and eq. $(3.66)$ says $$R^{\rho}_{\ \ \sigma\mu\nu}V^{\sigma}=[\nabla_{\mu},\nabla_{\nu}]V^{\rho}+T_{\mu\nu}^{\ \ \ \ \lambda}\nabla_{\lambda}V^{\rho}$$

To try and figure out how this came about, I started with the general coordinate-free definition of Riemann curvature tensor $$R(X,Y)V=\nabla_X\nabla_YV-\nabla_Y\nabla_XV-\nabla_{[X,Y]}V$$

Now I express LHS in component form and for the RHS, I use the identity $\nabla_X(\nabla_YV)=\nabla_{\nabla_XY}V+\nabla^2_{X,Y}V$ (see Lee, Introduction to Riemannian Manifolds proposition $4.21$)

$$\begin{align*}

X^{\mu}Y^{\nu}V^{\sigma}R^{\rho}_{\ \ \sigma\mu\nu}\partial_{\rho} &= \nabla_{\nabla_XY}V+\nabla^2_{X,Y}V-\nabla_{\nabla_YX}V-\nabla^2_{Y,X}V-\nabla_{[X,Y]}V

\\ &=\nabla_{\nabla_XY}V-\nabla_{\nabla_YX}V-\nabla_{[X,Y]}V+\nabla^2_{X,Y}V-\nabla^2_{Y,X}V

\\ &=\nabla_{T(X,Y)}V+\nabla^2_{X,Y}V-\nabla^2_{Y,X}V

\\ &=X^{\mu}Y^{\nu}T_{\mu\nu}^{\ \ \ \ \lambda}\nabla_{\partial_{\lambda}}V+\nabla^2_{X,Y}V-\nabla^2_{Y,X}V

\\ &=X^{\mu}Y^{\nu}T_{\mu\nu}^{\ \ \ \ \lambda}\nabla_{\partial_{\lambda}}V+\nabla^2V(\text{d}x^{\rho},Y,X)\partial_{\rho}-\nabla^2V(\text{d}x^{\rho},X,Y)\partial_{\rho}

\\ &=X^{\mu}Y^{\nu}T_{\mu\nu}^{\ \ \ \ \lambda}(\nabla_{\partial_{\lambda}}V)^{\rho}\partial_{\rho}+X^{\mu}Y^{\nu}\nabla^2V(\text{d}x^{\rho},\partial_{\nu},\partial_{\mu})\partial_{\rho}-X^{\mu}Y^{\nu}\nabla^2V(\text{d}x^{\rho},\partial_{\mu},\partial_{\nu})\partial_{\rho}

\\ &=X^{\mu}Y^{\nu}\big(T_{\mu\nu}^{\ \ \ \ \lambda}(\nabla_{\partial_{\lambda}}V)^{\rho}+\nabla^2V(\text{d}x^{\rho},\partial_{\nu},\partial_{\mu})-\nabla^2V(\text{d}x^{\rho},\partial_{\mu},\partial_{\nu})\big)\partial_{\rho}

\end{align*}$$

In the fifth equality I've used the definition of the second order covariant derivative: $\nabla^2_{X,Y}V(\ldots)=\nabla^2V(\ldots,Y,X)$. From the above, I get

$$X^{\mu}Y^{\nu}\big(V^{\sigma}R^{\rho}_{\ \ \sigma\mu\nu}\big)\partial_{\rho}=X^{\mu}Y^{\nu}\big(T_{\mu\nu}^{\ \ \ \ \lambda}(\nabla_{\partial_{\lambda}}V)^{\rho}+\nabla^2V(\text{d}x^{\rho},\partial_{\nu},\partial_{\mu})-\nabla^2V(\text{d}x^{\rho},\partial_{\mu},\partial_{\nu})\big)\partial_{\rho}$$

$$\implies V^{\sigma}R^{\rho}_{\ \ \sigma\mu\nu}=T_{\mu\nu}^{\ \ \ \ \lambda}(\nabla_{\partial_{\lambda}}V)^{\rho}+\nabla^2V(\text{d}x^{\rho},\partial_{\nu},\partial_{\mu})-\nabla^2V(\text{d}x^{\rho},\partial_{\mu},\partial_{\nu})$$

Now comparing this to the first equation (eq. $3.66$) in this post, I can make two conclusions:

1. The notation $\nabla_{\lambda}V^{\rho}$ actually denotes the $\rho$ component of $\nabla_{\lambda}V$, i.e. $\nabla_{\lambda}V^{\rho}\equiv(\nabla_{\partial_{\lambda}}V)^{\rho}$

2. The notation $\nabla_{\mu}\nabla_{\nu}V^{\rho}$ denotes the component of the tensor $\nabla\nabla V$, i.e., $\nabla_{\mu}\nabla_{\nu}V^{\rho}\equiv\nabla\nabla V(\text{d}x^{\rho},\partial_{\nu},\partial_{\mu})$

Even if $X,Y$ are taken to be coordinate basis vector fields (say $X=\partial_{\mu}$ and $Y=\partial_{\nu}$), sure the term involving the commutator involved in Riemann tensor's definition will vanish, but anyways that commutator term gets integrated into the expression for torsion tensor (see third equality in the calculation). So I think this general approach covers the case where $X,Y$ are basis vector fields (please correct me if I'm wrong).

My question is, are the above calculation and conclusions correct?