When decomposing $K_6$ into edge-disjoint partitions that are complete subgraphs we consider $K_5,\ldots K_2$. The $K_6$ as a subgraph of itself is trivial. I found that when we decompose a $K_6$ we have $8$ options: $15 K_2s$, $4K_3s + 3K_2s$, $3K_3s + 6K_2s$, $2K_3s + 9K_2s$, $1K_3 + 12K_2s$, $1K_4 + 1K_3 + 6K_2s$, $1K_4 + 9K_2s$, $1K_5 +5K_2s$. For some of them I have an explanation why we cannot have other forms, for example $1K_5$ leaves us with a star $1,5$ so the only option is to take $K_2s.$
What about $1K_4 + 1K_3 + 6K_2s$? Is there an easy way to explain these?
Let's talk about the case you mentioned, i.e. $1K_4$$+1K_3$$+6K_2$s. It is easy to see that upon removal of a $K_4$ from a complete hexagonal graph, you're left with no other $K_4$ and can at best get a $K_3$ (this is because there are only two vertices, say $v_1$ and $v_2$, that are connected to every other vertex so at best, they can be connected to the same vertex and make a triangle). One sees that all triangles share the edge $v_1$$v_2$ so if you remove any one triangle, you essentially remove the edge $v_1$$v_2$ and you can't have more triangles in the graph.