Any tree with $k$ edges $T$ can decompose any $2k$-regular graph $G$ into its copies I think this question is related to 'Any $2k$-regular graph can be $k$-factored', but I have no idea how to use this fact to solve the problem.
I also tried to use 'If $T$ is a tree with $k$ edges and $\delta(G)\geq k$, $T \subset G$'.
But I cannot repeat applying this to given $G$ since $\deg(v) \leq k$ for $v \in V(T)$ so that $\delta(G-T) \geq k$, and if I do it once more, $\delta(G-T-T') \geq 0$ for $T'$ the copy of $T$, then I cannot assure $G-T-T'$ contains $T$ as its subgraph.
The only thing I know is there should be one more condition: the girth of $G$ should be larger than $\mathrm{diam}(T)$.
Otherwise, $T$ cannot cover the smallest cycle in $G$.
Would you help me?
 A: 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.
A: Since I completed a more concrete solution, I post another answer.
Let $G$ be a graph with $\vert V(G) \vert = n$ and $T$ be a tree with $V(T)=\{v_1,\cdots,v_{m+1}\}$. 
Claim: $G$ is decomposed to $n$ copies of $T$, say $T_1,\cdots,T_n$, and for arbitrary $i$, the corresponding vertices to $v_i$ in $G$ are different among all copies of $T$.
Lemma: Every $2k$-regular graph is $2$-factorable.
We will use mathematical induction on $m$. 
For basis, when $m=1$, the statement is true since we can find a decomposition of a cycle (or a disjoint union of cycles) with $n$ edges satisfying the statement. 
Suppose the statement holds for all $m \leq k-1$.
Now consider $m=k$ case.
By the lemma, $2k$-regular graph $G$ has a $2$-factor. 
Delete one $2$-factor, say $C$, from $G$.
Now we get $2(k-1)$-regular graph.
By IH, this $2(k-1)$-regular graph is decomposed to $n$ copies of a tree $T'$ with $k-1$ edges, say $T_1',\cdots,T_n'$. 
WLOG assume $T'=T-v_{k+1}$ where $v_{k+1}$ is a leaf of $T$ adjacent to $v_k$. 
It suffices to show by adding $C$ again, we can construct $T$ from each of the copies of $T'$ where the corresponding vertices to $v_{k+1}$ in $G$ are not duplicated.
Here, observe that since all $n$ copies of $v_k$ in $T_1',\cdots,T_n'$ are different, so they cover all $n$ vertices of $G$. 
And also, $C$ covers all vertices of $G$.
Now give an orientation to $C$ 'consistently' so that each vertex has one in-coming edge and one out-going edge.
Then for each copy of $v_k$, there is exactly one out-going edge from $v_k$.
For each $T_j'$, select this out-going edge for the copy of $\{v_k,v_{k+1}\}$.
Now the copies of $v_{k+1}$ are not duplicated in $G$ by the construction of the orientation on $C$.
Also, $T_j' \cup \{v_k,v_{k+1}\}$ forms $T_j$. 
Moreover, since there are $n$ copies of $\{v_k,v_{k+1}\}$, we used all $n$ edges of $C$ - we covered entire $G$ by the copies of $T$. 
Note that $T_j$ in $G$ does not contain a cycle by the girth condition. $\square$
