TITLE: Leibniz rule for covariant derivative
QUESTION [7 upvotes]: 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.

REPLY [7 votes]: Say you have two tensors $\omega,\eta$ of valence $(0,1)$ (i.e., $1$-forms). Then $\nabla\omega$ and $\nabla\eta$ are $(0,2)$ tensors. For two vectors $u,v$, what should $\nabla\omega(u,v)$ mean? The usual thing, which you did, is to interpret it as $(\nabla_u\omega)(v)$, but someone may (although unlikely) interpret it as $(\nabla_v\omega)(u)$. This is not a problem, since almost everyone understands the first meaning and we are all happy.
Now, if $u,v,w$ are vectors, what do you think $\nabla(\omega\otimes\eta)(u,v,w)$ should be? of course, the standard answer is $\nabla_u(\omega\otimes\eta)(v,w)$, which equals $\nabla_u\omega(v)\eta(w)+\omega(v)\nabla_u\eta(w)$ (you can prove it).
However, notice what happens if you apply $\nabla\omega\otimes\eta+\omega\otimes\nabla\eta$ to the same tuple of vectors $(u,v,w)$ using the same convention: you get
$$\begin{align*}
(\nabla\omega\otimes\eta+\omega\otimes\nabla\eta)(u,v,w)
&= \nabla\omega(u,v)\eta(w)+\omega(u)\nabla\eta(v,w) \\
&= \nabla_u\omega(v)\eta(w)+\omega(u)\nabla_v\eta(w) \\
&\neq \nabla_u\omega(v)\eta(w)+\omega(v)\nabla_u\eta(w) \\
&= \nabla_u(\omega\otimes\eta)(v,w) \\
&= \nabla(\omega\otimes\eta)(u,v,w)
\end{align*}$$
which is not what we expected. That is why $\nabla(\omega\otimes\eta)\neq\nabla\omega\otimes\eta+\omega\otimes\nabla\eta$.
How to fix this? Well, let me advertise my second favourite option: using abstract index notation. In this convention, we use indices to indicate the slots of a tensor, and the tensor product is just juxtaposition. For example, the contraction of $\omega$ with a vector $v$ is written as $\omega(v)=\omega_av^a$, the tensor product $\omega\otimes\eta$ looks like $(\omega\otimes\eta)_{ab}=\omega_a\eta_b$, the covariant derivative (without being applied to a vector) is $(\nabla \omega)_{ab}=\nabla_a\omega_b$ and the applied covariant derivative is $(\nabla_u\omega)_a=u^b\nabla_b\omega_a$.
How does this help us? Well, then it is true that $(\nabla(\omega\otimes\eta))_{abc}=\nabla_a(\omega_b\eta_c)=\nabla_a\omega_b\eta_c+\omega_b\nabla_a\eta_c$. This is because the indices keep track of who should eat whom in the case three wilde vectors $u^av^bw^c$ appear. Don't let your notation to inconvenience you.
In this notation, your calculation is written as
$$\begin{align*}
\nabla_u(\nabla_vw)
&= u^a\nabla_a(v^b\nabla_bw^c) \\
&= u^a[\nabla_av^b\nabla_bw^c+v^b\nabla_a\nabla_bw^c] \\
&= u^a\nabla_av^b\nabla_bw^c+u^av^b\nabla_a\nabla_bw^c \text.
\end{align*}$$
Let me know if you have any question.