I've been learning about the covariant derivative and I have some doubts.
This answer suggests that $\nabla_{\mathbf{u}} T = \nabla T (\mathbf{u})$, where $T$ is a tensor. The tensor $\nabla T$ appears to be acting on the vector $\mathbf{u}$ in the same way a covector acts on a vector to give a scalar.
The answer then proceeds to derive the identity $\nabla^2_{\mathbf{u}, \mathbf{v}} = \nabla_{\mathbf{u}} \nabla_{\mathbf{v}} \mathbf{w} - \nabla_{\nabla_{\mathbf{u}} \mathbf{v}} \mathbf{w}$, where $\mathbf{u}$, $\mathbf{v}$ and $\mathbf{w}$ are vectors.
According to my interpretation, $$\nabla_{\mathbf{u}} \nabla_{\mathbf{v}} \mathbf{w} = \nabla_{\mathbf{u}} (\nabla \mathbf{w} (\mathbf{v})) \\ = \underbrace{(\nabla_{\mathbf{u}} (\nabla \mathbf{w}))}_{\text{a (1,1) tensor}} (\mathbf{v}) + \nabla \mathbf{w} (\nabla_{\mathbf{u}} \mathbf{v}) \\ = \underbrace{\nabla \nabla \mathbf{w}}_{\text{a (1,2) tensor}}(\mathbf{u}, \mathbf{v}) + \nabla_{\nabla_{\mathbf{u}} \mathbf{v}} \mathbf{w} \\ = \nabla^2_{\mathbf{u}, \mathbf{v}} + \nabla_{\nabla_{\mathbf{u}} \mathbf{v}} \mathbf{w} \\ \therefore \nabla^2_{\mathbf{u}, \mathbf{v}} = \nabla_{\mathbf{u}} \nabla_{\mathbf{v}} \mathbf{w} - \nabla_{\nabla_{\mathbf{u}} \mathbf{v}} \mathbf{w}.$$
My confusion arises here. Let $T$ and $S$ be tensors. The above derivation make use of some version of the Leibniz rule that appears to be of the form $\nabla_{\mathbf{u}}(T(S)) = (\nabla_{\mathbf{u}} T)(S) + T(\nabla_{\mathbf{u}} S)$. Is my interpretation correct?
Yet according to this answer, the rule $\nabla (T\otimes S) = \nabla T \otimes S + T\otimes \nabla S$ doesn't exist, but when you add a direction $\mathbf{u}$ $\nabla_{\mathbf{u}} (S\otimes T) = \nabla_\mathbf{u} S \otimes T + S \otimes \nabla_\mathbf{u} T$, it suddenly becomes true. Why?
I'm quite confused by these various versions of the Leibniz rule and the "total covariant derivative" $\nabla$ versus the covariant derivative $\nabla_{\mathbf{u}}$. I appreciate if someone could clear it up for me a little.