I've been asked to prove the following;
Let $G$ be a 4-regular graph, let $v$ be a vertex of $G$, and let $u_1, u_2, u_3, u_4$ be neighbours of $v$ in $G$. Replacing $v$ with a $K_4$ means adding four new vertices $v_1, v_2, v_3, v_4$, six new edges, $\{v_1, v_2 \}, \{v_1, v_3 \}, \{v_1, v_4 \}, \{v_2, v_3 \}, \{v_2, v_4 \}, \{v_3, v_4 \}$, and four other new edges, $\{u_1, v_1\}, \{u_2, v_2\}, \{u_3, v_3\}, \{u_4, v_4\}$, and deleting the vertex $v$ and the four edges $\{v, u_1\}, \{v, u_2\}, \{v, u_3\}, \{v, u_4\}$.
For any given 4-regular graph $G$, define the new graph $G^*$ to be the graph obtained from $G$ by replacing each vertex of $G$ with a $K_4$. Prove that $G$ has a Hamilton decomposition if and only if $G^*$ has a Hamilton decomposition.
For reference, here's what is meant, for the example of the $K_5$ graph;
Now, I am just trying to flesh out a decent argument for the statement "if $G$ has a Hamilton decomposition, then $G^*$ has a Hamilton decomposition", as I'm confident in proving the other direction if I can get this one correct.
So, to begin, we consider any 4-regular graph, which can be separated into two edge-disjoint cycles. In replacing every vertex of our 4-regular graph with a $K_4$ graph, we retain the edges that formed the Hamilton decompositions. Note that each $K_4$ graph can be split into two edge-disjoint Hamilton paths, which, when correctly integrated into the original edge sets forming the Hamilton decomposition of $G$, will form a Hamilton decomposition of $G^*$.
My problem with this is that I feel I haven't really defined explicitly the "correct integration" of the edge-disjoint $K_4$ graphs. For example, in writing out a few examples, I've found that the $K_4$ graphs have to be split up like this; 
With one of each pair being included in each Hamilton decomposition (as can be seen from the earlier picture). I just feel that a direct-proof such as this lacks definition, and I'd be interested to hear of any ways I could make this proof tighter.
Sorry about the length of this post, I just wanted to make sure it was clear what I was asking. :)