Let $\alpha$ be a differential form on a smooth manifold $M$. For simplicity, let's suppose that it is a $1$-form. Then we can think of $\alpha$ as a smooth map from $M$ to $T^* M$, the cotangent bundle. 

The exterior derivative $d\alpha$ is a $2$-form on $M$ that somehow "differentiates" $\alpha$. On the other hand, a $2$-form is a special kind of (i.e. alternating) bundle map from $TM$ to $T^* M$. Thinking of $\alpha$ as a smooth map, we obtain a map $D \alpha$ from $TM$ to $TT^* M$. Given a linear Ehresmann connection on the vector bundle $T^* M$, there is a linear map $\phi$ from $TT^* M$ to $T^* M$, with the latter viewed as the sub-bundle of $TT^* M$ which projects to $0$ in $M$. So, $\phi \circ D \alpha$ is a bundle map from $TM$ to $T^* M$, also known as a section of $T^*M \otimes T^* M$. This has a canonical projection $P$ to its skew-symmetrization $T^* M \wedge T^* M$, so $P \circ \Phi \circ D \alpha$ gives a section of $T^* M \wedge T^* M$, also known as a $2$-form. 

Perhaps more transparently, we can write this in terms of the covariant derivative associated to the Ehresmann connection. This is an operator $\nabla$ which maps sections $\alpha$ of $T^*M$ to a "$T^*M$-valued $1$-form" $\nabla \alpha: X \mapsto \nabla_X \alpha$, where $\nabla_X \alpha$ is a $1$-form, such that the map is tensorial in $X$ and follows the Leibniz rule in $\alpha$. This is related to $\phi$ precisely by the formula $\phi \circ D \alpha(X) = \nabla_X \alpha$, so this is the same as the previous definition, up to the skew-symmetrization. This should replace the contravariant $2$-tensor $\nabla \alpha(X, Y) = (\nabla_X \alpha)(Y)$ with the $2$-form $\omega_\alpha(X, Y) = (\nabla_X \alpha)(Y) - (\nabla_Y \alpha)(X)$. By the Leibniz rule for $\nabla$, we have \begin{align*}\omega_{f\alpha}(X, Y) &= \nabla_X (f\alpha)(Y) - \nabla_Y (f\alpha)(X)\\
&= f \omega_\alpha(X, Y) +  X(f)\alpha(Y) - Y(f) \alpha(X)\\ 
&= f\omega_\alpha(X, Y) + (df \wedge \alpha)(X, Y)
\end{align*}

So the map $\alpha \mapsto \omega_\alpha$ satisfies the usual derivation property of the exterior derivative and produces an honest $2$-form. *However*, it uses the entirely non-intrinsic Ehresmann connection $\nabla$. What gives? 

Does something like this work for higher $k$-forms? Replacing the bundle $T^* M$ with $\wedge^k T^* M$ everywhere in the arguments gives an analogous definition of an "exterior derivative" of an arbitrary $k$-form, but it is not so clear that this now satisfies $d(\omega_1 \wedge \omega_2) = d\omega_1 \wedge \omega_2 + (-1)^k \omega_1 \wedge d \omega_2$. 


**EDIT** I realized this was confusing - when I say an Ehresmann connection, I'm referring to one that is linear, so it is equivalent to the usual notion of covariant derivative for a vector bundle. However, I wanted the definition to include connections that aren't necessarily determined by an affine connection on $TM$.