A vector field $\vv$ can be thought of as a vector-valued 0-form. As such, we should be able to take its exterior derivative and obtain a vector-valued 1-form $d\vv$.

The simplest examples occur when $\vv$ can be expanded in terms of a rectangular basis, e.g. \begin{equation} \FF = F^x \,\xhat + F^y \,\yhat \end{equation} in two dimensions. Surely, since $\xhat$ and $\yhat$ are constant vector fields, we should have \begin{equation} d\xhat = 0 = d\yhat \end{equation} so that \begin{equation} d\FF = dF^x \,\xhat + dF^y \,\yhat \end{equation} In an arbitrary basis, we would have \begin{equation} \FF = F^i \, \ee_i \end{equation} and now \begin{equation} d\FF = dF^i \, \ee_i + F^i \, d\ee_i \end{equation} Thus, to determine the exterior derivative of vector fields, we need to know how to take the exterior derivative of our basis vectors.