Distribution of descents of random permutations

Yes. Put $V_i = U_1+U_2+\cdots+U_i$, $W_i = \lfloor V_i \rfloor$ and $X_i = V_i - W_i$.

I claim that the $X_i$ are independent and uniformly distributed in $[0,1]$. Indeed, conditional on $(U_1, \ldots, U_{i-1})$, we have $V_i = U_i + c$ for some constant $c$, so $V_i$ is (conditional on $(U_1, \ldots, U_{i-1})$) uniformly distributed in $[c, c+1]$, and then $X_i$ is uniform in $[0,1]$.

In particular, with probability $1$, the $X_i$ are distinct, so there is a unique order preserving map $\{ X_1, \ldots, X_n \} \longrightarrow \{ 1, 2, \ldots, n \}$. Let $\sigma$ be the permutation such that $X_i \mapsto \sigma(i)$. So $\sigma$ is uniform in $S_n$.

Then $\sigma(i) < \sigma(i+1)$ iff $W_i = W_{i-1}$, and $\sigma(i) > \sigma(i+1)$ iff $W_i = W_{i-1}+1$. So the total number of descents of $\sigma$ is $\sum_{i=1}^n (W_i - W_{i-1}) = W_n - W_0 = W_n$, as you requested.

This is a pretty standard construction. I think of it as coming from Stanley, "Eulerian partitions of a unit hypercube", 1977, although your citation is from 1973, so it was before Stanley!