Perhaps I would rephrase the first dot point as: For $m=3$ the 3-simplices of the 2-coskeleton correspond to all possible choices of four triangles in $X$ which can be arranged into a tetrahedral shape.$^\dagger$ and finish at the end with: $\dagger$ I.e. to a map on the $2$-truncations $(\Delta^3)^{(2)}\to X_\bullet^{(2)}$, which is determined by where the $4$ non-degenerate $2$-simplicies are mapped, such that these form a tetrahedron.