First part of MTW Exercise 10.1: additivity of the commutator. Is there a vector symbolic proof that doesn't use components? The image shows vector fields $\vec{x},\vec{y},\vec{a}=\vec{x}+\vec{y},\vec{b}.$  The small black arrows are $\left[\vec{x},\vec{b}\right],\left[\vec{y},\vec{b}\right],\left[\vec{a},\vec{b}\right].$  The goal is to show
$$\left[\vec{x},\vec{b}\right]+\left[\vec{y},\vec{b}\right]
=\left[\vec{x}+\vec{y},\vec{b}\right]
=\left[\vec{a},\vec{b}\right].$$

I believe that what MTW call the "commutator" and "closer of quadrilaterals" is identical to what is also known as Lie brackets.  In all of what follows, I am assuming a coordinate induced basis.
My question addresses the first part of MTW Exercise 10.1.  I give a component based solution below.  But I sense that there is an abstract, conceptual aspect of the exercise that I am missing.
Is there a way to establish the additivity of the commutator using symbolic tensor (vector) notation, without appeal to component formulations?  I am specifically addressing the first step in the exercise where we are asked to show:
$$\left[\mathbf{u},\mathbf{v}+\mathbf{w}\right]=\left[\mathbf{u},\mathbf{v}\right]+\left[\mathbf{u},\mathbf{w}\right]$$
The essence of my component-based proof (below) is that the second derivative terms cancel.  In terms of a coordinate induced basis, this is equivalent to saying that the commutators of the basis vectors vanish.

Exercise 10.1 ADDITIVITY OF COVARIANT DIFFERENTIATION
Show that the commutator ("closer of quadrilaterals") is additive
$$
\left[\mathbf{u},\mathbf{v}+\mathbf{w}\right]=\left[\mathbf{u},\mathbf{v}\right]+\left[\mathbf{u},\mathbf{w}\right];\\
\left[\mathbf{u}+\mathbf{n},\mathbf{v}\right]=\left[\mathbf{u},\mathbf{v}\right]+\left[\mathbf{n},\mathbf{v}\right],
$$
Use this result, the additivity condition
$$
\nabla_{\mathbf{u}}\left(\mathbf{v}+\mathbf{w}\right)=\nabla_{\mathbf{u}}\mathbf{v}+\nabla_{\mathbf{u}}\mathbf{w},
$$
and the symmetry of the covariant derivative,
$$
\nabla_{\mathbf{u}}\mathbf{v}-\nabla_{\mathbf{v}}\mathbf{u}=\left[\mathbf{u},\mathbf{v}\right],
$$
to prove that
$$
\nabla_{\mathbf{u}+\mathbf{n}}\mathbf{v}=\nabla_{\mathbf{u}}\mathbf{v}+\nabla_{\mathbf{n}}\mathbf{v}.
$$



Box 10.2 E



E. In the real physical world, be it Newtonian or relativistic, parallel
transport of a triangle cannot break its legs apart: (1) $\mathbf{A},\mathbf{B},\mathbf{C}$ initially
such that $\mathbf{A}+\mathbf{B}=\mathbf{C};$ (2) $\mathbf{A},\mathbf{B},\mathbf{C}$ each
parallel transported with himself by freely falling (inertial) observer;
(3) then $\mathbf{A}+\mathbf{B}=\mathbf{C}$ always. Any other result
would violate the equivalence principle!

*

*Consequence of this (as seen by following through definition of
covariant derivative, and by noting that any vector $\mathbf{u}$
can be regarded as the tangent vector to a freely falling world line):
$$
\mathbf{\nabla}_{\mathbf{u}}\left(\mathbf{v}+\mathbf{w}\right)=\mathbf{\nabla}_{\mathbf{u}}\mathbf{v}+\mathbf{\nabla}_{\mathbf{u}}\mathbf{w}
$$
for any vector $\mathbf{u}$ and vector fields $\mathbf{v}$ and $\mathbf{w}.$

*Consequence of this, combined with symmetry of covariant derivative,
and with additivity of the closer of quadrilaterals $\left[\mathbf{u},\mathbf{v}\right]:$
$$
\nabla_{\mathbf{u}+\mathbf{n}}\mathbf{v}=\nabla_{\mathbf{u}}\mathbf{v}+\mathbf{\nabla}_{\mathbf{n}}\mathbf{v}.
$$
(See exercise 10.1.) This can be inferred, alternatively, from the
equivalence principle: in a local inertial frame, as in special relativity
or Newtonian theory, the change in $\mathbf{v}$ along $\mathbf{u}+\mathbf{n}$ should
equal the sum of the changes along $\mathbf{u}$ and along $\mathbf{n}.$

The covariant derivative along a curve with tangent vector $d/d\lambda=\mathbf{u}$ as it appears in Box 10.2 B is

$$
\nabla_{\mathbf{u}}\mathbf{v}=\lim_{\epsilon\to0}\left\{ \frac{\left[\mathbf{v}\left(\lambda_{0}+\epsilon\right)\right]_{\text{parallel transported to}\lambda_{0}}-\mathbf{v}\left(\lambda_{0}\right)}{\epsilon}\right\} .
$$

The commutator defined by equation 9.19 where $\mathbf{u},\mathbf{v}$ are vector fields, and $f$ is a scalar field is

$$\left[\mathbf{u},\mathbf{v}\right]f=\mathbf{u}\left\{ \mathbf{v}\left[f\right]\right\} -\mathbf{v}\left\{ \mathbf{u}\left[f\right]\right\}$$

From this we can derive a component based expression for the commutator operator, $\left[\mathbf{u},\mathbf{v}\right]$
$$\begin{aligned}
\left[\mathbf{u},\mathbf{v}\right]f
&=\left(\mathit{f}_{,\alpha\beta}v^{\alpha}
+\mathit{f}_{,\alpha}v^{\alpha}{}_{,\beta}\right)u^{\beta}
-\left(\mathit{f}_{,\alpha\beta}u^{\alpha}
+\mathit{f}_{,\alpha}u^{\alpha}{}_{,\beta}\right)v^{\beta}\\
&=\mathit{f}_{,\alpha}\left(v^{\alpha}{}_{,\beta}u^{\beta}
-u^{\alpha}{}_{,\beta}v^{\beta}\right)\\
&=\mathbf{e}_{\alpha}\left[f\right]\left(v^{\alpha}{}_{,\beta}u^{\beta}
-u^{\alpha}{}_{,\beta}v^{\beta}\right)\\
\therefore\left[\mathbf{u},\mathbf{v}\right]
&=\mathbf{e}_{\alpha}\left(v^{\alpha}{}_{,\beta}u^{\beta}
-u^{\alpha}{}_{,\beta}v^{\beta}\right)
\end{aligned}.$$
At this point, since only components and their partial derivatives appear in the expression, we can substitute $\mathbf{v}\rightarrow\mathbf{v}+\mathbf{w}$ to establish the first form
$$\left[\mathbf{u},\mathbf{v}+\mathbf{w}\right]=\left[\mathbf{u},\mathbf{v}\right]+\left[\mathbf{u},\mathbf{w}\right].$$
The second form follows trivially from the anti-symmetry inherent in the definition of the commutator.
Using my interpretation of "the covariant derivative of $\mathbf{v}$ along $\mathbf{u}$," (Box 10.2 B.) where $\mathbf{u}=d/d\lambda$
$$
\frac{d\mathbf{v}}{d\lambda}
=\nabla_{\mathbf{u}}\mathbf{v}
=\frac{\partial\mathbf{e}_{\alpha}v^{\alpha}}{\partial x^{\mu}}u^{\mu},
$$
the commutator operator can be expressed in therms of directional derivatives of vector fields, without reference to a scalar field
$$\begin{aligned}\left[\mathbf{u},\mathbf{v}\right] 
& =\nabla_{\mathbf{u}}\mathbf{v}-\nabla_{\mathbf{v}}\mathbf{u}\\
& =\left(\mathbf{e}_{\nu}v^{\nu}\right)_{,\mu}u^{\mu}- 
   \left(\mathbf{e}_{\mu}u^{\mu}\right)_{,\nu}v^{\nu}\\
 & =\left[\mathbf{e}_{\mu},\mathbf{e}_{\nu}\right]v^{\nu}u^{\mu}
+\mathbf{e}_{\alpha}\left(v^{\alpha}{}_{,\mu}u^{\mu}
-u^{\alpha}{}_{,\nu}v^{\nu}\right)\\
 & =\mathbf{e}_{\alpha}\left(v^{\alpha}{}_{,\mu}u^{\mu}-u^{\alpha}{}_{,\nu}v^{\nu}\right)
\end{aligned}.
$$
But all of this relies on the use of component formulations.  The ordering of statements suggests to me that the additivity of the commutator is to be established without appeal to "the additivity for covariant differentiation" appearing in Box 10.2 E.
$\color{red}{\text{Note well:}}$ The following needs more scrutiny!
"Composition" could mean either $\partial_{\mathbf{u}}\left(\partial_{\mathbf{v}}f\right)=\nabla_{\mathbf{u}}\nabla_{\mathbf{v}}f=\mathbf{u}\left[\mathbf{v}\left[f\right]\right],$ or $\mathbf{w}\left[f\right]=\left(\nabla_{\mathbf{u}}\mathbf{v}\right)\left[f\right]=\nabla_{\mathbf{w}}f.$  They are not identical.  Both are linear with constant coefficients, but only the latter follows the Leibniz rule, and thus qualifies as a vector field.  Their differences cancel out when the order of $\mathbf{u}$ and $\mathbf{v}$ are reversed and subtracted from the respective original expressions.  That is "the commutator nullifies their differences". But my argument is very sloppy, and needs refinement. 
 

I'm adding this here rather than posting it as an answer, because I don't believe it follows the logical sequence intended in the exercise.
Following an approach suggested in the comments, it appears that we
have all of the following, for an arbitrary scalar field
$f$, constants $\mathrm{a},\mathrm{b}\in\mathbb{R},$ and vector
fields $\mathbf{u},\mathbf{v},\mathbf{w}:$
The vector fields form a vector space
$$
\mathbf{u}\left(\mathrm{a}f\right)+\mathbf{v}\left(\mathrm{b}f\right)=\mathrm{a}\mathbf{u}\left(f\right)+\mathrm{b}\mathbf{v}\left(f\right)=\left[\mathrm{a}\mathbf{u}+\mathrm{b}\mathbf{v}\right]\left(f\right).
$$
Since the vector fields map the set of differentiable scalar fields
into itself
$$
\mathbf{u},\mathbf{v}:\mathscr{C}^{\infty}\left(M\right)\to\mathscr{C}^{\infty}\left(M\right),
$$
the composition of vector fields does likewise
$$
\mathbf{u}\left(\mathbf{v}\right):\mathscr{C}^{\infty}\left(M\right)\to\mathscr{C}^{\infty}\left(M\right).
$$
Thus the composition of vector fields is additive in its argument
$$
\mathbf{u}\left(\mathrm{a}\mathbf{v}+\mathrm{b}\mathbf{w}\right)=\mathrm{a}\mathbf{u}\left(\mathbf{v}\right)+\mathrm{b}\mathbf{u}\left(\mathbf{w}\right).
$$
This can be written in the form used in the exercise
$$
\nabla_{\mathbf{u}}\left(\mathbf{v}+\mathbf{w}\right)=\nabla_{\mathbf{u}}\mathbf{v}+\nabla_{\mathbf{u}}\mathbf{w}.
$$
Since the vector fields form a vector space, we have the final result
to be found in the exercise
$$
\nabla_{\left(\mathbf{u}+\mathbf{v}\right)}\mathbf{w}=\nabla_{\mathbf{u}}\mathbf{w}+\nabla_{\mathbf{v}}\mathbf{w}.
$$
From this, the additivity of the commutator follows easily.  See https://physics.stackexchange.com/a/696984/117014
 A: Indeed, there is a short proof that uses only the abstract properties of the quantities you mentioned. I'll sketch it and let you fill in the details.
Consider the commutators $[\mathbf{u},\mathbf{v}] = \nabla_{\mathbf{u}} \mathbf{v} - \nabla_{\mathbf{v}} \mathbf{u}$ and $[\mathbf{n},\mathbf{v}] = \nabla_{\mathbf{n}} \mathbf{v} - \nabla_{\mathbf{v}} \mathbf{n}$. Adding them leads to $[\mathbf{u}+\mathbf{n},\mathbf{v}]$, which we know satisfies $[\mathbf{u}+\mathbf{n},\mathbf{v}] = \nabla_{\mathbf{u} + \mathbf{n}} \mathbf{v} - \nabla_{\mathbf{v}} (\mathbf{u} + \mathbf{n})$ from the veery definition of the commutator. Just open up the expression $[\mathbf{u},\mathbf{v}] + [\mathbf{n},\mathbf{v}] = [\mathbf{u}+\mathbf{n},\mathbf{v}]$ and you should get the result.

As noticed in the comments,
$$[\mathbf{u},\mathbf{v}](f) = \mathbf{u}(\mathbf{v}(f)) - \mathbf{v}(\mathbf{u}(f))$$
is bilinear. This can be derived from the fact that vector fields are linear. For example,
\begin{align}
[\mathbf{u} + \mathbf{n},\mathbf{v}](f) &= (\mathbf{u} + \mathbf{n})(\mathbf{v}(f)) - \mathbf{v}((\mathbf{u} + \mathbf{n})(f)), \\
&= \mathbf{u}(\mathbf{v}(f)) + \mathbf{n}(\mathbf{v}(f)) - \mathbf{v}(\mathbf{u}(f) + \mathbf{n}(f)), \\
&= \mathbf{u}(\mathbf{v}(f)) + \mathbf{n}(\mathbf{v}(f)) - \mathbf{v}(\mathbf{u}(f)) - \mathbf{v}(\mathbf{n}(f)), \\
&= \mathbf{u}(\mathbf{v}(f)) - \mathbf{v}(\mathbf{u}(f)) + \mathbf{n}(\mathbf{v}(f)) - \mathbf{v}(\mathbf{n}(f)), \\
&= [\mathbf{u},\mathbf{v}](f) + [\mathbf{n},\mathbf{v}](f).
\end{align}
A: 
Is thete a way to establish the additivity of the commutator using symbolic tensor (vector) notation without components?

Yes there is.
However, using the covariant derivative is not the right way to go about it. This is because the Lie bracket, what you call a commutator, doesn't depend on a connection and hence doesn't depend on the covariant derivative. Recall, given a metric we can construct a torsion free connection that is compatible with the metric called the Levi-Civita connection and then we build the covariant derivative out of this. The Lie bracket just depends on the smooth structure of the manifold and not its geometry, aka the metric.
First, let us fix a smooth manifold $M$. Let $XM$ denote the space of tangent fields on $M$. Then:

$XM \simeq Drv(CM)$

Here, $Drv(CM)$ is the space of derivations on the space of smooth functions, $CM$. The above isomorphism is demonstrated in Baez & Munions, Knots & Quantum Gravity and they explain what is meant by a derivation. Now, any space of derivations is closed under the commutator and this turns it into a Lie algebra. Pullback this Lie algebra to $XM$ and this gives the Lie bracket on tangent fields. Since the bracket is bilinear we are done.
In more detail, Let

$\sharp: Drv(CM) \rightarrow XM$


and $\flat: XM \rightarrow Drv(CM)$

exhibit the isomorphism explained above.
We define:

$u.f := u^{\flat}.f$ for $u \in XM$ and $f \in CM$.

And we also define:

$[u,v]:= [u^{\flat}, v^{\flat}]^{\sharp}$

This is the 'pullback' mentioned above. Then:

$[u,v].f$


$= [u^{\flat},Y^{\flat}]^{\sharp\flat}.f$


$= [u^{\flat}, v^{\flat}].f$


$= (u^{\flat}.v^{\flat}.f - v^{\flat}.u^{\flat}.f)$


$= u.(v.f) - v.(u.f)$

This is the usual formula for the Lie bracket.
And:

$[u,v+w]$


$= [u^{\flat}, (v+w)^{\flat}]^{\sharp}$


$= [u^{\flat}, v^{\flat} + w^{\flat}]^{\sharp}$


$= [u^{\flat}, v^{\flat}]^{\sharp} + [u^{\flat},w^{\flat}]^{\sharp}$
                     


$= [u,v] + [u,w]$

which is the additivity formula you were looking for. And notice - no covariant derivative, connection or metric was used!
A: This is just a first stab at an answer.  Any help refining it will be appreciated.
Dr. Wheeler tried to explained this to me decades ago.  It finally came back to me.  At the time, the explanation baffled me.  Now it makes a little bit of sense.
It goes something like this:
Let $\mathbf{v}$ and $\mathbf{w}$ be small relative to $\mathbf{x}$ and $\mathbf{u};$ where $\mathbf{0}= \left[\mathbf{u},\mathbf{x}\right].$  This means $\left[\mathbf{u},\mathbf{x}\right]=\left[\mathbf{u},\mathbf{0}\right],$ and that $\left[\mathbf{u},\mathbf{x}+\mathbf{v}\right]= \left[\mathbf{u},\mathbf{0}+\mathbf{v}\right].$
Since $\mathbf{v}$ is small relative to $\mathbf{x},$ adding another small vector $\mathbf{w}$ to $\mathbf{x}+\mathbf{v}$ will displace $\left[\mathbf{u},\mathbf{x}+\mathbf{v}\right]$ by approximately the same amount as adding $\mathbf{w}$ to $\mathbf{x}$ will displace $\left[\mathbf{u},\mathbf{x}\right];$  that is, by $\left[\mathbf{u},\mathbf{w}\right].$
This is just a hand-wavy outline:
\begin{align*}
\mathbf{0}= & \left[\mathbf{u},\mathbf{x}\right]=\left[\mathbf{u},\mathbf{0}\right]\\
\left[\mathbf{u},\mathbf{x}+\mathbf{v}\right]= & \left[\mathbf{u},\mathbf{0}+\mathbf{v}\right]\\
= & \left[\mathbf{u},\mathbf{x}\right]+\left[\mathbf{u},\mathbf{v}\right]\\
\mathbf{x}+\mathbf{v}= & \mathbf{x}+\mathcal{O}\left[\mathbf{v}\right]\\
\left[\mathbf{u},\mathbf{v}+\mathbf{w}\right]= & \left[\mathbf{u},\mathbf{x}+\mathbf{v}+\mathbf{w}\right]\\
= & \left[\mathbf{u},\mathbf{x}+\mathcal{O}\left[\mathbf{v}\right]+\mathbf{w}\right]\\
= & \left[\mathbf{u},\mathbf{x}+\mathcal{O}\left[\mathbf{v}\right]\right]+\left[\mathbf{u},\mathbf{w}\right]\\
= & \left[\mathbf{u},\mathbf{x}+\mathbf{v}\right]+\left[\mathbf{u},\mathbf{w}\right]\\
= & \left[\mathbf{u},\mathbf{v}\right]+\left[\mathbf{u},\mathbf{w}\right]
\end{align*}
