You're right that things change for m>1; I was thinking sloppily.
Assume $U=\{1,\ldots,n\}$ for concreteness. If $Y_1,\ldots,Y_m$ are chosen independently and uniformly from $U$, then for any $k_1,\ldots,k_m\in U$, we of course have
$$
\Pr[Y_1=k_1,\ldots,Y_m=k_m] = \frac{1}{n^m}.
$$
On the other hand, if $x=(x_1,\ldots,x_m)$ is chosen uniformly from the standard $n$-simplex and $Y_1,\ldots,Y_m$ are then chosen independently according to $x$, then
$$
\Pr[Y_1=k_1,\ldots,Y_m=k_m] = \mathbb{E}\Pr[Y_1=k_1,\ldots,Y_m=k_m|x]
= \mathbb{E}\prod_{i=1}^m x_{k_i} = \frac{n!}{(n+r)!}\prod_{j=1}^m frac{n!}{(n+r)!}\prod_{j=1}^n r_j!,
$$
where $r_j = \#\{i\in U #\{1\le i \le m : k_i=j\}$ and $r=r_1 + \cdots r_m$r_n$. This last expectation can be proved most easily from Lemma 1 in this paper.
