Different approaches to $g\left(\nabla_{\nabla_{e_j}e_i}X,e_i\right)+g\left(\nabla_{e_i}X,\nabla_{e_j}e_i\right)=0$. 
Let $S$ be a $(1,1)$-tensor. Let $(e_i)$ be a local frame and let $(e^i)$ be its dual. Prove that
$$X\left(S\left(e_i,e^i\right)\right)=(\nabla_XS)\left(e_i,e^i\right).\tag{1}$$

This formula exactly means that $\nabla_X$ commutes with the contraction. We can prove this via normal coordinates:
Let $(x^i)$ be normal coordinates centered at $p$. Then
\begin{align}
\left(\nabla_{\frac{\partial}{\partial x^j}}S\right)\left(\frac{\partial}{\partial x^i},dx^i\right)(p)
&=
\frac{\partial}{\partial x^j}\Big|_p\left(S\left(\frac{\partial}{\partial x^i},dx^i\right)\right)
-S\left(\nabla_{\frac{\partial}{\partial x^j}\big|_p}\frac{\partial}{\partial x^i},dx^i\right)
-S\left(\frac{\partial}{\partial x^i},\nabla_{\frac{\partial}{\partial x^j}\big|_p}dx^i\right)\\
&=\frac{\partial}{\partial x^j}\left(S\left(\frac{\partial}{\partial x^i},dx^i\right)\right)(p)
\end{align}
Since formula (1) is cleary coordinate-independent, the conclusion easily follows.
Now I want to learn more approaches other than the above one.
In other words, I am wondering other natural methods to show
$$S\left(\nabla_Xe_i,e^i\right)+S\left(e_i,\nabla_Xe^i\right)=0,\tag{2}$$
where (2) is clearly equivalent to (1).
Any helps would be highly appreciated!
Added
Maybe the original problem seems trivial. Then I want to ask another problem:

Let $(e_i)$ be a local orthonormal frame. Prove that
$$g\left(\nabla_{\nabla_{e_j}e_i}X,e_i\right)+g\left(\nabla_{e_i}X,\nabla_{e_j}e_i\right)=0$$
for any vector field $X$.

This is clearly a corollary of the original problem. Now I am wondering other approaches. Thanks for in advance for any hints!
 A: Here is a generalization using the language of bundles, maybe it is helpfull now or in the future (or never :P), atleast it is a different approach.
Let $M$ be a smooth manifold. In general if $E,F$ are bundles with connection over $M$ one can define a connection on the bundle $\mathrm{Hom}(E,F)$ by requiring the product rule $\nabla (Ae)=(\nabla A)e+A(\nabla e)$ for all $A\in\Gamma\mathrm{Hom}(E,F)$, $e\in\Gamma E$. The trivial bundle $M\times\mathbb R$ has a natural connection $d$.
Therefore given $E$  with connection one has connections on $E^*$, $\mathrm {End}E$, $(\mathrm {End}E)^*$ and so on. Then  the trace $\mathrm{tr}$ is a section of $(\mathrm {End}E)^*$  The claim now is that $\nabla \mathrm{tr}=0$ or equivalently $(\nabla \mathrm{tr})S=0$ for all $S\in\Gamma\mathrm {End}E$. It suffices to show this for $S$ of the form $S=e^*\otimes e$ with $e\in\Gamma E$,  $e^*\in\Gamma E^*$. Notice $\mathrm{tr}(e^*\otimes e)=e^*(e)$, so
$$
d(\mathrm{tr}S)
=d(e^*(e))
=(\nabla e^*)(e)+e^*(\nabla e)\\
=\mathrm{tr}((\nabla e^*)\otimes e+e^*\otimes\nabla e)
=\mathrm{tr}(\nabla(e^*\otimes e))
=\mathrm{tr}(\nabla S)
$$
Here also the product rule for the tensor product was used. Hence $(\nabla\mathrm{tr})S=d(\mathrm{tr}S)-\mathrm{tr}(\nabla S)=0$ and so $\nabla\mathrm{tr}=0$.
A: Here is another answer to your original question:
First, let's recall that given a connection $\nabla$ on a vector bundle $E$, there are uniquely defined connections on $E$ and $E^*\otimes E$, where for any sections $v$ of $E$ and $\theta$ of $E^*$,
\begin{align*}
  \partial_X\langle\theta,v\rangle &= \langle\nabla_X\theta,v\rangle + \langle \theta,\nabla_Xv\rangle\\
  \nabla_X(\theta\otimes v) &= (\nabla_X\theta)\otimes v + \theta\otimes\nabla_Xv.
\end{align*}
Let $(e_1, \dots, e_m)$ be a frame of $E$ with dual frame $(e^1, \dots, e^m)$. We want to show that given any section $S$ of $E^*\otimes E$,
$$
  S(\nabla_{X}e_i, e^i) + S(e_i,\nabla_X e^i) = 0.
$$
It suffices to prove it for $S = \theta\otimes v$. This follows by the following straightforward calculation:
\begin{align*}
  (\theta\otimes v)(\nabla_{X}e_i, e^i) + (\theta\otimes v)(e_i,\nabla_X e^i)
  &= \langle\theta,\nabla_Xe_i\rangle\langle e^i,v\rangle
    +
    \langle \theta,e_i\rangle\langle \nabla_Xe^i,v\rangle\\
  &= (\partial_X\langle\theta,e_i\rangle
    - \langle\nabla_X\theta, e_i\rangle)\langle e^i,v\rangle\\
&\quad
    + \langle\theta,e_i\rangle(\partial_X\langle e^i,v\rangle - \langle e^i,\nabla_Xv\rangle)\\
  &= \partial_X(\langle\theta,e_i\rangle\langle e^i,v\rangle)
    - \langle \nabla_X\theta, e_i\rangle\langle e^i,v\rangle
    - \langle \theta,e_i\rangle\langle e^i,\nabla_Xv\rangle\\
  &= \partial_X\langle\theta,v\rangle - \langle\nabla_X\theta,v\rangle - \langle\theta,\nabla_Xv\rangle\\
  &= 0
\end{align*}
