Covariant derivative of a covariant derivative I'm trying to find the covariant derivative of a covariant derivative, i.e. $\nabla_a (\nabla_b V_c)$. 
This is something I've taken for granted a lot in calculations, namely I though that by the Leibniz rule we just have:
$$\nabla_a (\nabla_b V_c) = \partial_a(\nabla_b V_c) - \Gamma_{ab}^{d}\nabla_c V_d - \Gamma_{ac}^{d} \nabla_d V_c$$
However when we prove that the covariant derivative of a $(0,2)$ tensor is the above, we use the fact that the covariant derivative satisfies a Leibniz rule on $(0,1)$ tensors: $\nabla_a(w_b v_c) = v_c\nabla_a(w_b) + w_b\nabla_a(v_c)$. However $\nabla_a$ on it's own is not a tensor so how do we have the above formula for it's covariant derivative?
 A: The term $\nabla_b V_c$ is a (0,2) tensor writing in the abstract index notation,  when writing in full basis form it reads
\begin{equation} 
\nabla_b V_c \;dx^b\otimes dx^c\;,
\end{equation}
Now the status of $\nabla_b V_c $ is a components it is a scalar function while $dx^b\otimes dx^c$ is a basis of (0,2)-tensor.
Then the double covariant derivative reads
\begin{equation} 
\nabla \Big(  \nabla_b V_c \;dx^b\otimes dx^c  \Big)\;, 
\end{equation}
where $$\nabla_b V_c \equiv \partial_b V_c -\Gamma_b{}^q{}_c V_q\;.$$
The Leibniz rule is needed in this step
\begin{eqnarray}
\nabla \Big(  \nabla_b V_c \;dx^b\otimes dx^c  \Big)&=& \nabla \Big(  \nabla_b V_c \Big)\;dx^b\otimes dx^c  + \nabla_b V_c \;\nabla \Big( dx^b \Big)\otimes dx^c + \nabla_b V_c \;dx^b\otimes \nabla \Big( dx^c  \Big)\;,\\
&=& \nabla_m \Big( \overbrace{ \nabla_b V_c}^{a\; scalar} \Big)\;dx^m\otimes dx^b\otimes dx^c  + \nabla_b V_c \;\times \Big(-\Gamma^b{}_n dx^n \Big)\otimes dx^c \\&&+ \nabla_b V_c \;dx^b\otimes  \times \Big( -\Gamma^c{}_p dx^p \Big)\;,\\
&=& \nabla_m \Big( \overbrace{ \nabla_b V_c}^{a\; scalar} \Big)\;dx^m\otimes dx^b\otimes dx^c  -\Gamma^b{}_n \nabla_b V_c \;  dx^n \otimes dx^c \\&& -\Gamma^c{}_p \nabla_b V_c \;dx^b\otimes   dx^p \;,\\
&=&\partial_m \Big( \overbrace{ \nabla_b V_c}^{a\; scalar} \Big)\;dx^m\otimes dx^b\otimes dx^c  -\Gamma_r{}^b{}_n dx^{r} \otimes \nabla_b V_c \; dx^{n} \otimes dx^c \\&&-\Gamma_s{}^c{}_p dx^{s} \otimes \nabla_b V_c \;dx^b  \otimes dx^{p} \;,\\
&=&\partial_m \Big(  \nabla_b V_c \Big)\;dx^m\otimes dx^b\otimes dx^c  - \Gamma_r{}^b{}_n\nabla_b V_c \;  dx^r \otimes dx^n \otimes dx^c \\&&-\Gamma_s{}^c{}_p \nabla_b V_c \;dx^s\otimes   dx^b\otimes dx^p \;,\\
&=&\Big[\partial_m \Big(  \nabla_b V_c \Big)- \Gamma_m{}^d{}_b\nabla_d V_c-\Gamma_m{}^e{}_c \nabla_b V_e\Big ]\;dx^m\otimes dx^b\otimes dx^c \;.
\end{eqnarray}
Then we define
$$
\nabla \Big(  \nabla_b V_c \;dx^b\otimes dx^c  \Big)=:\nabla_m \nabla_b V_c \;dx^m \otimes dx^b\otimes dx^c
$$
Finally, in abstract index notation we have
$$
\nabla_m \nabla_b V_c \equiv \partial_m \Big(  \nabla_b V_c \Big)- \Gamma_m{}^d{}_b\nabla_d V_c-\Gamma_m{}^e{}_c \nabla_b V_e
$$
A: Easy way
Let me first state the straight-forward way to do this computation.
$$
\langle \nabla_a \nabla_b V, \partial_c\rangle =
\partial_a \langle \nabla_b V, \partial_c \rangle - \langle \nabla_aV, \nabla_a \partial_c\rangle = \partial_a (\nabla_bV)_c - (\nabla_bV)_d \Gamma_{ac}^d
$$
First equality follows from compatibility, second equality uses definition of Levi-Civita symbols.
Hard way
You are suggesting a roundabout way to do this, which formalizes to the following: 
$$
\nabla_a\nabla_bV = \nabla_a\left[~(\nabla_cV\otimes dx^c)[\partial_b]~\right]
= \nabla_a\left[~C(\nabla_cV\otimes dx^c \otimes \partial_b)~\right]
= C [\nabla_a (\nabla_cV\otimes dx^c \otimes \partial_b)]
$$
where $$
C: T_pM \otimes T_pM \otimes T^*_pM \to T_pM, ~~ w\otimes z\otimes V \to
z[V]w
$$
is the contraction map of the last two arguments. Covariant derivative on mixed-type tensors commute with contractions (used in the last equality). Observe the expression within $C[ \cdots ]$ is a covariant derivative of a mixed tensor, which you can compute with the Leibneiz rule, and use your favorite component-wise formulas.
