The usual definition does not include $0$ in the second factor, the reason being that $(M\times[0,\infty)/\sim,dt^2+t^2g)$ is not smooth at the origin unless $M$ is a round sphere (here $\sim$ identifies $M\times\{0\}$ to a point). I have edited the original answer to reflect that the origin is missing. Also, the metric cone is always noncompact, because of the second factor.

I usually see $R(X,Y,W,Z)$ instead of $R(W,Z,X,Y)$. One reason to prefer this is seen in Penrose's abstract index notation, where $$ \nabla_i \nabla_j Z^k - \nabla_j \nabla_i Z^k = R_{ij}{}^k{}_l Z^l . $$ This one naturally expresses the Riemann curvature tensor as an $\mathrm{End}(TM)$-valued two-form and also preserves the order of the indices. If you lower the index, this recovers your display as $R(X,Y,W,Z)$.

Oops, Willie is of course correct. My answer has been fixed.

I edited the answer to include a reference to a more complete answer to your first question. For your new question in the comments, $D(1)=0$ for any Riemannian manifold and any vector field.

Note that he is taking the limit as $s\to0$. The basic idea is that in geodesic normal coordinates, $g_{ij}=\delta_{ij}+O(s^2)$ for $s$ small, and hence $V(p,s)=\omega_ns^n + O(s^{n+2})$. In your example of hyperbolic space, $V(p,s)=\omega_n\int_0^s\sinh^{n-1}t\,dt$ and the claimed limit follows from the fact $s^{-1}\sinh s\to1$ as $s\to0$.

Consider the difference of the left-hand side and the first summand. This consists of terms where at least one derivative hits $v/u$. The second summand is when all derivatives hit $v/u$, and the final one is the cross term when one derivative from $\Delta$ hits $u$ and the other hits $v/u$.