Here's an answer via a summation, but not a closed form solution.
First, lets rewrite the setting more clearly:
Let $G$ be a ground set, with $|G|=n$ elements. You prefer to call them $x_1, x_2, \dots, x_n$.
Your "set 1" is equivalent to the set of non-empty subsets of $G$. Normally the set of all subsets of $G$ (aka power set of $G$) would be denoted $2^G$ or $\mathcal{P}(G)$ (with a "fancy" $\mathcal{P}$ or $\mathbb{P}$), but you want to exclude the empty subset case. I will arbitrarily call it:
$$\text{Your set 1} = Q(G) = \{ S \subset G: S \neq \emptyset\}$$
Each element $S \in Q(G)$ is a non-empty subset of $G$, say $S = \{x_1, x_3, x_7\}$. Instead of using set-theoretic notation you prefer to write it as a linear combination, e.g. $S = x_1 + x_3 + x_7$, but these are clearly equivalent to subsets - assuming the standard convention that $+$ is commutative (i.e. ordering doesn't matter). Also, we have $|Q(G)| =2^n - 1$.
Given any $G$, define $H(G)$ to be the set of all unordered pairs of elements of $G$:
$$H(G) = \{ \{x_i, x_j\} : x_i \in G, x_j \in G, x_i \neq x_j \}$$
Instead of writing each pair in the set-theoretic way $\{x_i, x_j\}$ you prefer to write it as a product $x_i x_j$. Again these are clearly equivalent to unordered pairs - assuming your multiplication is also commutative, i.e. $x_i x_j = x_j x_i$. Also, we have $|H(G)| = {n \choose 2} = n(n-1)/2$.
Your "set 2" is simply the set of non-empty subsets of $H(G)$:
$$\text{Your set 2} = Q(H(G)) = \{T \subset H(G): T \neq \emptyset\}$$
Again, every element of $T \in Q(H(G))$ is a non-empty set consisting of pairs, and you prefer to write it as a linear combination: e.g. $T = \{ \{x_1, x_3\}, \{x_1, x_7\} \} = x_1 x_3 + x_1 x_7$. Also we have $|Q(H(G))| = 2^{|H(G)|} - 1 = 2^{n(n-1)/2} - 1$.
Then you want to count the number of "allowed" pairs of the form $(S, T)$ where:
$S \in Q(G)$ i.e. your set 1
$T \in Q(H(G))$ i.e. your set 2
For every pair $\{a, b\} \in T$, we require $a \in S, b\in S$.
- This requirement rules out e.g. $x_1 + x_2 + x_3 + x_1 x_4$ because $x_4$ appears in a pair but not as a "singleton".
There is no requirement that the "other direction" is satisfied, i.e. there can be $x \in S$ which appears in no pairs $\in T$. E.g. $x_1 + x_2 + x_3 + x_1 x_2$ is allowed (and must be counted) even though $x_3$ does not appear in any pair.
Now, my answer:
Consider a subset $S \subset G$ of size $k$. For each such subset, you can form ${k \choose 2} = k(k-1)/2$ pairs from the elements of $S$, and any non-empty collection of such pairs is allowed. I.e. the number of allowed pairs of $(S,T)$ for this specific $S$, is
$$2^{k(k-1)/2} - 1$$
Now of course there are ${n \choose k}$ subsets of $G$ of size $k$, so you just need to sum over them:
$$\text{total no. of allowed pairs} = f(n) = \sum_{k=2}^n {n \choose k} (2^{k(k-1)/2} - 1)$$
Sorry I don't know how to simplify this further. Anyway, the values starting from $f(2)=1, f(3)=10$ are:
$$1, 10, 97, 1418, 40005, 2350474, 286192257 \dots$$
and I don't think this is in OEIS
Best Answer
We can write $C_n$, for every non-negative integer $n$, as $$C_n = \sum_{(s_1, s_2, \dots, s_n) \in \{-1, 1\}^n}s_1s_2\cdots s_n \cdot f\Big(s_1x_1 + s_2x_2 + \dots + s_nx_n\Big),$$ where $\{-1, 1\}^n := \Big\{(s_1, s_2, \dots, s_n) \mid s_i \in \{-1, 1\} \text{ for all positive integers } i \leq n\Big\}$.
The idea is to just use the "variables" $s_i$ to act as the signs.
For example, $\{-1, 1\}^3$ is defined to be $$\{(\color{red}{-1}, \color{red}{-1}, \color{red}{-1}), (\color{red}{-1}, \color{red}{-1}, \color{green}{+1}), (\color{red}{-1}, \color{green}{+1}, \color{red}{-1}), (\color{red}{-1}, \color{green}{+1}, \color{green}{+1}), (\color{green}{+1}, \color{red}{-1}, \color{red}{-1}), (\color{green}{+1}, \color{red}{-1}, \color{green}{+1}), (\color{green}{+1}, \color{green}{+1}, \color{red}{-1}), (\color{green}{+1}, \color{green}{+1}, \color{green}{+1})\},$$ so we have \begin{align*} C_3 = & \: (\color{red}{-1})(\color{red}{-1})(\color{red}{-1}) \cdot f\Big((\color{red}{-1})x_1 + (\color{red}{-1})x_2 + (\color{red}{-1})x_3\Big) \:+ \\ & \: (\color{red}{-1})(\color{red}{-1})(\color{green}{+1}) \cdot f\Big((\color{red}{-1})x_1 + (\color{red}{-1})x_2 + (\color{green}{+1})x_3\Big) \:+ \\ & \: (\color{red}{-1})(\color{green}{+1})(\color{red}{-1}) \cdot f\Big((\color{red}{-1})x_1 + (\color{green}{+1})x_2 + (\color{red}{-1})x_3\Big) \:+ \\ & \: (\color{red}{-1})(\color{green}{+1})(\color{green}{+1}) \cdot f\Big((\color{red}{-1})x_1 + (\color{green}{+1})x_2 + (\color{green}{+1})x_3\Big) \:+ \\ & \: (\color{green}{+1})(\color{red}{-1})(\color{red}{-1}) \cdot f\Big((\color{green}{+1})x_1 + (\color{red}{-1})x_2 + (\color{red}{-1})x_3\Big) \:+ \\ & \: (\color{green}{+1})(\color{red}{-1})(\color{green}{+1}) \cdot f\Big((\color{green}{+1})x_1 + (\color{red}{-1})x_2 + (\color{green}{+1})x_3\Big) \:+ \\ & \: (\color{green}{+1})(\color{green}{+1})(\color{red}{-1}) \cdot f\Big((\color{green}{+1})x_1 + (\color{green}{+1})x_2 + (\color{red}{-1})x_3\Big) \:+ \\ & \: (\color{green}{+1})(\color{green}{+1})(\color{green}{+1}) \cdot f\Big((\color{green}{+1})x_1 + (\color{green}{+1})x_2 + (\color{green}{+1})x_3\Big). \end{align*}