7. For any $n$-element set $S_{n}$, if its subsets $A_{1}$, $A_{2}, \cdots, A_{k}$ satisfy $\bigcup_{i=1}^{k} A_{i}=S_{n}$, then the unordered set group $\left(A_{1}, A_{2}, \cdots, A_{k}\right)$ is called a “$k$-stage partition” of the set $S_{n}$. Therefore, the number of 2-stage partitions of $S_{n}$ is $\qquad$
7. $\frac{1}{2}\left(3^{n}+1\right)$.

Let $F_{n}$ be the number of second-order partitions of an $n$-element set $S_{n}=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$.
Obviously, $F_{0}=1, F_{1}=2$.
Suppose $\left(A_{i}, B_{i}\right)\left(i=1,2, \cdots, F_{n}\right)$ is any partition of $S_{n}$. To obtain the partition of $S_{n+1}=S_{n} \cup\left\{a_{n+1}\right\}$, we only need to add $a_{n+1}$ to $A_{i}$ and $B_{i}$, which results in three types:
(1) $\left(A_{i} \cup\left\{a_{n+1}\right\}, B_{i}\right)$;
(2) $\left(A_{i}, B_{i} \cup\left\{a_{n+1}\right\}\right)$;
(3) $\left(A_{i} \cup\left\{a_{n+1}\right\}, B_{i} \cup\left\{a_{n+1}\right\}\right)$.
Also, note that when $A_{i}=B_{i}$,
$$
\left(A_{i} \cup\left\{a_{n+1}\right\}, B_{i}\right) \text { and }\left(A_{i}, B_{i} \cup\left\{a_{n+1}\right\}\right)
$$

are the same, at this time, $S_{n}=A_{i} \cup B_{i}=A_{i}=B_{i}$. It is only at this time that a pair is repeated.
$$
\begin{array}{l}
\text { Hence } F_{n}=3 F_{n-1}-1=3\left(3 F_{n-2}-1\right)-1 \\
=3^{n} F_{0}-\sum_{k=0}^{n-1} 3^{k}=\frac{1}{2}\left(3^{n}+1\right) .
\end{array}
$$