Any tree with $k$ edges $T$ can decompose any $2k$-regular graph $G$ into its copies

Edit: It seems that my proof works only if the girth is strictly larger than the diameter of $T$.

Color each one of the $k$ edges of the tree with a different color, and give it an arbitrary orientation.
Every $2k$-regular graph is $2$ factorable, so we can give an edge coloration with $k$ colors. On each cycle of the factorization, orient all the edges consistently. That way, each vertex has one incoming and one outgoing edge, of each color.

Take an edge of the graph, and find its corresponding edge in the tree. From it, add incrementally edges of the graph according to the tree. At each edge, as every type of edge is present in every direction, it will always be possible to build the tree. The subgraph so constructed has no cycle due to the girth condition. Thus the subgraph is a tree. By construction, we see that, for each edge, the corresponding tree is uniquely defined. So we can repeat the process and cover the entire graph.