Exterior derivative well defined on vector fields Let $M$ be a manifold and $d$ the exterior derivative on forms. We extend $d$ to vector fields as follows: if $X = \sum_i X_i \partial_i$ in local coordinates, then
$$dX = \sum_i (dX_i) \partial_i$$
i.e. if $Y$ is another vector field then $dX(Y)$ is the vector field $\sum_i (dX_i)(Y) \partial_i$. I am trying to show that this is well defined, independent of coordinate changes. When I calculate it out explicitly I get a rather nasty expression which does not seem to simplify. A coordinate-free definition of $d$ would be ideal, but I would also be satisfied with a coordinate based calculation. 

This operation is not well defined.  One way to see this is to note that, according to your definition, a vector field $X$ satisfies $dX=0$ if and only if it is a linear combination of the coordinate vector fields $\partial_i$ with constant coefficients (at least when the coordinate domain is connected).  When you change to a different coordinate system, the same vector fields will typically not have constant coefficients.  
The map $X\mapsto dX$ you've defined is an example of a connection.  Given any smooth local frame for $TM$ on some subset $U\subset M$ (for example, a coordinate frame), there is a unique connection defined on $U$ for which the basis vector fields are parallel; but the connection depends heavily on the choice of frame.
