Given a (smooth) vector bundle $E\to M$ with connection $\nabla$ one can define the exterior derivative $d^\nabla:\Omega^k(M,E)\to\Omega^{k+1}(M,E)$ by $$d^\nabla\omega(X_0,\dots,X_k):=\sum_{i=0}^k(-1)^k\nabla_{X_i}\bigl(\omega(X_0,\dots,\widehat{X_i},\dots,X_k)\bigr)+\sum_{0\leq i<j\leq k}(-1)^{i+j}\omega([X_i,X_j],X_0,\dots,\widehat{X_i},\dots,\widehat{X_j},\dots,X_k)$$ for all vector fields $X_0,\dots,X_k\in\Gamma TM$, where $[X,Y]h=X(Yh)-Y(Xh)$ is the Lie bracket, and hats denote omission. However, to me this formula looks pretty random, not at all natural like many other things in differential geometry. My question now is whether there is some intuitive way to derive this formula?
For example, to define the derivative $f'$ of a function $f:\mathbb R\to\mathbb R$ I first start with a description (or set of axioms) of what $f'$ is supposed to do, i.e. "assign every $x\in\mathbb R$ the rate of change of $f$ at $x$", then notice that $f$ changes by $f(\tilde x)-f(x)$ over $[x,\tilde x]$, so a good approximation of its rate of change would be $\frac{f(\tilde x)-f(x)}{\tilde x-x}$, which becomes more accurate as $\tilde x$ goes to $x$, so I define $f'(x)=\lim_{\tilde x\to x}\frac{f(\tilde x)-f(x)}{\tilde x-x}$.
Is there a similar derivation of $d^\nabla$? One that starts with some (intuitive of rigorous) requirements that describe what $d^\nabla$ is supposed to do, and that arrives at the formula above?