Are vertex colourings--in a sense--edge colourings?
By analogy with the line graph, I think you are asking the following:


  For every undirected graph $G = (V,E)$, does there exist an undirected graph $G' = (V',E')$ such that each vertex $v \in V$ corresponds to an edge $(v_1,v_2) \in E'$ and the edges corresponding to $u \in V$ and $v \in V$ share at least one endpoint if and only if $(u,v) \in E$?


The answer can be seen to be no. Consider the four-vertex tree $G$ with root $v$ having three children $x,y,z$. In $G'$, we must have four edges: $(v_1,v_2),(x_1,x_2),(y_1,y_2),(z_1,z_2)$. Further, it must be the case that either $v_1$ or $v_2$ is an endpoint of each of the other three edges (i.e., $\left| \{v_1,v_2\} \cap \{x_1,x_2\} \right| \geq 1$, etc). But this means that at least two of the other three edges must share a common endpoint, which violates our requirements since no two of $x,y,z$ are adjacent in the original graph.

I think the same graph will give you a counterexample for the matching question as well.