(This post is based on my answer to a question posted on math.stackexchange.com.)

The exterior derivative of a $1$-form is usually defined as follows: Let $df$ denote the differential of a scalar function $f$. The exterior derivative is defined to be a linear operator from $1$-forms to $2$-forms that satisfies the following properties: Given any scalar function $f$ and $1$-form $\theta$, \begin{align*} d(df) &= 0\\ d(f\theta) &= df\wedge\theta + fd\theta. \end{align*} It is not difficult to show that, if such an operator exists, it is unique. We can use local coordinates to show that operator $d$ does exist locally by defining the exterior derivative of $\theta = a_i\,dx^i$ to be $$ d\theta = da_i\wedge\,dx^i. $$ It is straightforward to verify that this satisfies the properties above. Uniqueness of $d$ implies that it is in fact a global operator on the manifold.

The standard coordinate-free definition is, given vector fields $V$ and $W$, \begin{equation}\label{standard} \langle d\theta, V\otimes W\rangle = \langle V, d\langle W,\theta\rangle\rangle - \langle W, d\langle V,\theta\rangle\rangle - \langle\theta,[V,W]\rangle. \end{equation} It is necessary, however, to verify that $d\theta$ is in fact a tensor.

Here is another way that is new to me. Let $\nabla$ be a torsion-free connection. In other words, it defines a directional derivative of vector fields that satisfies the following properties: Given vector fields $U$, $V$, and $W$ and a scalar function $f$, \begin{align} \nabla_{U+V}W &= \nabla_UW + \nabla_VW \label{additive}\\ \nabla_{fV}W &= f\nabla_VW\\ \nabla_U(V+W) &= \nabla_UV + \nabla_UW\\ \nabla_V(fW) &= \langle V,df\rangle W + f\nabla_VW \label{derivation}\\ \nabla_VW - \nabla_WV &= [V,W] \label{torsion-free} \end{align} Equations \eqref{additive}-\eqref{derivation} define a connection, and equation \eqref{torsion-free} is the torsion-free property.

The connection can be extended uniquely to an operator on $1$-forms by assuming the product rule \begin{equation}\label{product} \langle W, d\langle \theta, V\rangle\rangle = \langle \nabla_W\theta,V\rangle + \langle \theta, \nabla_WV\rangle. \end{equation} Since the left side does not depend on the connection, neither does the right. Therefore, if $\tilde\nabla$ is another torsion-free connection, then $$ \langle \tilde\nabla_W\theta,V\rangle + \langle \theta, \tilde\nabla_WV\rangle = \langle \nabla_W\theta,V\rangle + \langle \theta, \nabla_WV\rangle $$ By this and the torsion-free property \eqref{torsion-free}, \begin{equation} \begin{split} \langle V, \tilde\nabla_W\theta\rangle - \langle W,\tilde\nabla_V\theta\rangle &= \langle V, \nabla_W\theta\rangle - \langle W,\nabla_V\theta\rangle + \langle\theta, \tilde\nabla_VW - \tilde\nabla_WV - \nabla_VW + \nabla_WV\rangle\\ &= \langle V, \nabla_W\theta\rangle - \langle W,\nabla_V\theta\rangle + \langle \theta, [V,W] - [V,W]\rangle\\ &= \langle V, \nabla_W\theta\rangle - \langle W,\nabla_V\theta\rangle. \end{split}\label{independent} \end{equation}

We now define the exterior derivative of $\theta$ to be the $2$-form $d\theta$ such that, if $V, W \in T_pM$, then $$ \langle d\theta(p), V\otimes W\rangle = \langle V, \nabla_W\theta(p)\rangle - \langle W,\nabla_V\theta(p)\rangle. $$ It follows directly from this definition that $d\theta(p)$ is an antisymmetric bilinear function on $T_pM$ and therefore a well-defined exterior $2$-tensor. By \eqref{independent}, this definition is independent of the connection used.

The standard coordinate-free definition \eqref{standard} can be obtained from this definition and \eqref{product}: \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*}