Exterior Derivative of One-Form vs Torsion of Connection Let $\omega$ be a $1$-form. Then $d\omega$ may be defined by the formula
$$
d\omega(X,Y) = \frac{\partial}{\partial X}\iota_Y\omega - \frac{\partial}{\partial Y}\iota_X\omega-\omega([X,Y])
$$
where $X,Y$ are vector fields. This formula bears a resemblance to the formula for the torsion of a connection $\nabla$:
$$
\nabla_XY - \nabla_YX - [X,Y]
$$
Is there a geometric explanation for this resemblance?
 A: I don't have an answer to your question, but it leads to a simple coordinate-free definition of the exterior derivative of a $1$-form $\theta$:
Let $\nabla$ be a torsion-free connection. It defines the covariant derivative of a vector field. The covariant derivative of a $1$-form is uniquely determined by the product rule
$$
d\langle \theta, V\rangle = \langle \nabla\theta, V\rangle + \langle \theta,\nabla V\rangle.
$$
Since the left side of the equation above does not depend on the connecton, it follows that if $\tilde\nabla$ is another torsion-free connection, then
\begin{equation}\label{change}
\langle \tilde\nabla\theta, V\rangle + \langle \theta,\tilde\nabla V\rangle = \langle \nabla\theta, V\rangle + \langle \theta,\nabla V\rangle.
\end{equation}
The exterior derivative of $\theta$ can be defined by
$$
\langle d\theta, V\otimes W\rangle
= \langle \nabla_V\theta,W\rangle - \langle V,\nabla_W\theta\rangle,
$$
where $V, W$ are tangent vectors at a point.
This definition makes it obvious that $d\theta$ is a well-defined exterior $2$-tensor. Using the equations above and the torsion-free property, it is easy to show that this definition does not depend on the connection.
The better known coördinate-free formula follows from the equations above and the torsion-free property, because
\begin{align*}
\langle d\theta, V\otimes W\rangle
&= \langle \nabla_V\theta,W\rangle - \langle V,\nabla_W\theta\rangle\\
&= \langle V,d\langle\theta,W\rangle\rangle - \langle \theta, \nabla_VW\rangle
- \langle W,d\langle\theta,V\rangle\rangle + \langle \theta,\nabla_WV\rangle\\
&= \langle V,d\langle\theta,W\rangle\rangle
- \langle W,d\langle\theta,V\rangle\rangle - \langle \theta,[V,W]\rangle\\
\end{align*}
If you choose local coordinates and use the flat connection with respect to those coordinates, then you get the usual formula for the exterior derivative.
A: Note a very deep result, but I found a common pattern and that the third term cancels some terms.
You gave two expressions and I found a third on Wikipedia:
$$\begin{align}
d\omega(X,Y) &= \partial_X \iota_Y \omega - \partial_Y \iota_X \omega - \omega([X,Y]) 
\\
T(X,Y) &= \nabla_X Y - \nabla_Y X - [X,Y]
\\
R(X,Y) &= \nabla_X \nabla_Y - \nabla_Y \nabla_X - \nabla_{[X,Y]}
\end{align}$$
They all have a common form,
$$
\Omega(X,Y) = D(X) A(Y) - D(Y) A(X) - A([X,Y]),
$$
where $D(X)=\partial_X+\Gamma_X$ for some automorphism $\Gamma_X$ (possibly vanishing). Below, $D_j=\partial_j+\Gamma_j$ will be used for $D(\partial_j).$ Also, $A_j=A(\partial_j).$
For the three cases we have
$$\begin{align}
d\omega(X,Y) &: D(X)=\partial_X,\ A(X)=\iota_X\omega = \omega(X)
\\
T(X,Y) &: D(X)=\nabla_X,\ A(X)=X
\\
R(X,Y) &: D(X)=\nabla_X,\ A(X)=\nabla_X
\end{align}$$
With $X=X^j\partial_j,\ Y=Y^k\partial_k$ we have
$$\begin{align}
D(X) A(Y)
&= D(X^j\partial_j) A(Y^k\partial_k)
= X^j D(\partial_j) (Y^k A(\partial_k))
\\
&= X^j (\partial_j+\Gamma_j) (Y^k A_k) \\
&= X^j ((\partial_j Y^k) A_k + Y^k (\partial_j A_k) + Y^k \Gamma_j A_k) \\
&= X^j (\partial_j Y^k) A_k + X^j Y^k (D_j A_k)
\\
D(Y) A(X) 
&= Y^k (\partial_k X^j) A_j + X^j Y^k (D_k A_j)
\\
A([X,Y]) &= A(X^j(\partial_j Y^k)\partial_k - Y^k(\partial_k X^j)\partial_j)
\\
&= X^j(\partial_j Y^k) A_k - Y^k (\partial_k X^j) A_j
\end{align}$$
and when we then simplify $\Omega(X,Y)$ we see that $A([X,Y])$ cancels two terms from $D(X) A(Y) - D(Y) A(X)$:
$$
D(X) A(Y) - D(Y) A(X) - A([X,Y])
\\
= \left( X^j (\partial_j Y^k) A_k + X^j Y^k (D_j A_k) \right)
- \left( Y^k (\partial_k X^j) A_j + X^j Y^k (D_k A_j) \right)
- \left( X^j(\partial_j Y^k) A_k + Y^k (\partial_k X^j) A_j \right)
\\= X^j Y^k (D_j A_k - D_k A_j)
$$

EDIT 15 July 2021
A deeper explanation is that all three of these are outer derivatives.
We are used to handle scalar-valued forms, but what if the form is vector-valued? Then we use the exterior covariant derivative, which I assume (without thorough investigation) can be written
$$
D\omega(X,Y) = \nabla_X\omega(Y) - \nabla_Y\omega(X) - \omega([X,Y]).
$$
Taking $\omega(X)=X$ makes $D\omega$ be the torsion, and taking $\omega_Z(X)=\nabla_X Z$ makes $D\omega$ be the curvature.
A: An underlying geometric relation is that the condition that $\nabla$ be torsion-free is precisely the condition that its action on differential forms reduces to the exterior derivative upon antisymmetrization. In particular, the tensorial Leibiz rule for $\nabla$ together with invariance under contraction yield the general coordinate-free formula for the action of $\nabla$ on a $(0,k)$-tensor $\omega$:
\begin{align}  
(\nabla \omega)(X_0, X_1, \dots, X_k) & = (\nabla_{X_0} \omega)(X_1, \dots, X_k) 
\\ & =  X_0 \left( \omega(X_1,\dots,X_k) \right) - \omega(\nabla_{X_0} X_1, \dots, X_k) - \cdots - \omega(X_1, \dots, \nabla_{X_0} X_k)
\end{align}
Meanwhile, if we take $\omega$ to be a differential $k$-form, Cartan's magic formula can be used to show that we may obtain a similar coordinate-free formula for the action of the exterior derivative:
$$(d \omega)(X_0, X_1, \dots, X_k) = \sum_{i=0}^k (-1)^i X_i \left( \omega(X_0,\dots, \widehat{X_i}, \dots,X_k) \right) + \sum_{i<j} (-1)^{i+j} \omega([X_i,X_j],X_0,\dots, \widehat{X_i}, \dots, \widehat{X_j}, \dots,X_k),$$
where hats denote omission. I'll leave it up to the reader to convince themselves that, if $\nabla$ is torsion-free, then antisymmetrizing the action of $\nabla$ on a $k$-form yields the action of $d$ (up to a factor of $k!$).
For example, in your $1$-form case, the former is
$$(\nabla \omega)(X,Y) = X(\omega(Y)) - \omega(\nabla_X Y) $$
Antisymmetrizing yields
$$\text{Alt}(\nabla \omega)(X,Y) = (\nabla \omega)(X,Y) - (\nabla \omega)(Y,X) = X(\omega(Y)) - Y(\omega(X)) - \omega( \nabla_X Y - \nabla_Y X)$$
If $\nabla$ is torsion-free, this is equivalent to your formula
$$(d \omega)(X,Y) = X(\omega(Y)) - Y(\omega(X)) - \omega([X, Y]). $$
