$\newcommand{\iii}{\mathbf i}
\newcommand{\K}{\mathcal K}$
Quoting a comment by the OP of the quoted question:
I am seeking such an explicit expression, but have not been able to correctly perform all the combinatoral accounting required. Ideally the final expression should not involve a summation over elements of $S$, instead summing over a certain dimensional partitioning in the sense of my previous comment.
Let us provide such an expression (one may assume here, slightly more generally, that $U$ is Lebesgue measurable). With $[n]:=\{1,\dots,n\}$, we have
\begin{equation*}
|U\cap S^k|=\sum_{(i_1,\dots,i_k)\in[n]^k}1_{(X_{i_1},\dots,X_{i_k})\in U},
\end{equation*}
where $X_1,\dots,X_n$ are iid points each uniformly distributed over the rectangle $R$ in $\mathbb R^d$.
So,
\begin{equation*}
E|U\cap S^k|=\sum_{(i_1,\dots,i_k)\in[n]^k}P\big((X_{i_1},\dots,X_{i_k})\in U\big). \tag{1}
\end{equation*}
The probability $P\big((X_{i_1},\dots,X_{i_k})\in U\big)$ depends on $\iii:=(i_1,\dots,i_k)$ only through the partition
\begin{equation*}
\Pi(\iii):=\{J_1(\iii),\dots,J_n(\iii)\}
\end{equation*}
of the set $[k]$,
where
\begin{equation*}
J_i(\iii):=\{j\in[k]\colon i_j=i\}
\end{equation*}
for $i\in[n]$. Take any such partition $\Pi(\iii)$ and clean it up by removing its empty elements and ordering the remaining, non-empty elements of it so that for the resulting ordered partition
\begin{equation*}
K=(K_1,\dots,K_s)
\end{equation*}
of the set $[k]$ we have the following: $s\in[k]$, the $K_\alpha$'s are pairwise disjoint, their union is $[k]$, and $\min K_\alpha<\min K_\beta$ whenever $1\le\alpha<\beta\le k$.
For any given $s\in[k]$, let $\K_s$ denote the set of all such ordered partitions $K$.
Also let $K(\iii)$ denote the ordered partition in $\K_s$ (for some $s=s(\iii)\in[k]$ depending on $\iii$) obtained from the $k$-tuple $\iii$ by the just described procedure.
Then to restore any $k$-tuple $\iii$ given the corresponding ordered partition $K(\iii)$, we only need to "color" the $s$ members of the partition $K(\iii)$ by $s$ different "colors" selected from the $n$ available "colors" $1,\dots,n$; there are
\begin{equation*}
n_{(s)}:=n(n-1)\dots(n-(s-1))
\end{equation*}
choices of such colorings of the given ordered partition $K=(K_1,\dots,K_s)$.
E.g., if $n=5$, $k=6$, and $\iii=(3,1,4,1,3,3)$, then only $s(\iii)=3$ different "colors" (namely, $1,3,4$) out of the $5$ available "colors" $1,\dots,5$ were actually used for this $\iii$, and the ordered partition $K(\iii)$ here is $K=(\{1,5,6\},\{2,4\},\{3\})$, where $\{1,5,6\}$ is the set of the positions of the $3$'s in this $\iii$, $\{2,4\}$ is the set of the positions of the $1$'s, and $\{3\}$ is the set of the positions of the $4$'s. The $k$-tuple $\iii=(3,1,4,1,3,3)$ is only one feasible coloring of this ordered partition $K=(\{1,5,6\},\{2,4\},\{3\})$; another feasible coloring is e.g. $(2,5,3,5,2,2)$.
However, given any such ordered partition $K\in\K_s$ with $s\in[k]$, the probability $P\big((X_{i_1},\dots,X_{i_k})\in U\big)$ will not further depend on the choice of $\iii=(i_1,\dots,i_k)$ with $K(\iii)=K$; this follows because the $X_i$'s are iid. In fact, for any such $K=(K_1,\dots,K_s)$ and any $\iii=(i_1,\dots,i_k)$ with $K(\iii)=K$
\begin{equation*}
P\big((X_{i_1},\dots,X_{i_k})\in U\big)=P\big((Y_1(K),\dots,Y_k(K))\in U\big),
\end{equation*}
where
\begin{equation*}
Y_j(K):=X_\alpha\quad\text{if}\quad j\in K_\alpha
\end{equation*}
for some $\alpha\in[s]$.
Thus, by (1),
\begin{equation*}
E|U\cap S^k|=\sum_{s=1}^k n_{(s)}
\sum_{K\in\K_s}P\big((Y_1(K),\dots,Y_k(K))\in U\big).
\end{equation*}
In particular, if $U=T_1\times\cdots\times T_k$, where $T_1,\dots,T_k$ are Lebesgue-measurable subsets of rectangle $R$, then
\begin{equation*}
E|U\cap S^k|=\sum_{s=1}^k \frac{n_{(s)}}{\lambda_d(R)^s}\,
\sum_{K\in\K_s}\prod_{\alpha=1}^s \lambda_d\Big(\bigcap_{j\in K_s}T_j\Big),
\end{equation*}
where $\lambda_d$ is the Lebesgue measure over $\mathbb R^d$.
