Equivalent formulation of $\sum_{A \subseteq X} \sum_{x\neq y, x,y \in A} f(x)f(y)$? For any finite set $X$, for all $f: X \longrightarrow \mathbb R$,
$$ \sum_{A \subseteq X} \sum_{x \neq y x,y \in A} f(x)\cdot f(y) = \sum_{A \subseteq X} \sum_{x \in A, y \not \in A} f(x) \cdot f(y).$$
I can show this equality by taking a bit of a detour, as follows. 
Let $N = |X|$. Fix an enumeration $\{A_i\}$ of all subsets of $X$ so that for each $i,j$, $|i - j| = 2^{N-1}$ iff  $A_j = X \setminus A_i$. For each $i \leq 2^{N}$, let $i^*$ be the unique $j$ such that $|i - j| = 2^{N-1}$. For each $i \leq 2^N$ and each $y \in A_i$, let $i_y$ be the unique $j$ such that $A_j = A_i \setminus \{y\}$.
Now, we can think of each term in the sum on the left hand side as an element of $${\cal A}:= \bigcup_i \{ ((x,i),(y,i)) : x \neq y, x,y \in A_i \}.$$
And we can think of each term in the sum on the right hand side as an element of $${\cal B}:= \bigcup_{i} \{ ((x,i),(y,i^*)) : x \in A_i, y \in A_{i^*} \}.$$
Note now that the following function is a bijection from $\cal A$ to $ \cal B$: $$\phi(((x,i),(y,i))) = ((x,i_y),(y,i_y^*)).$$
And this, I take it, establishes the desired result. But I'm hoping for a more straightforward proof. 
 A: Treat the two sides separately.  Grouping together all like terms on the left side, you can see that
$$\sum_{A \subseteq X} \sum_{\substack{x,y \in A
\\\\x \neq y}} f(x)\,f(y) = \sum_{\substack{x,y \in X
\\\\x \neq y}} m_{x,y} \,f(x)\, f(y)$$
where
$$m_{x,y} = \operatorname{card}\,\lbrace A \subseteq X \mid x\in A \mathrm{~and~} y\in A\rbrace.$$
Similarly, grouping together all like terms on the right side yields
$$\sum_{A \subseteq X}\sum_{x \in A, y \not \in A} f(x) \, f(y) =\!\!\sum_{\substack{x,y \in X
\\\\x \neq y}} n_{x,y} \,f(x)\, f(y)$$
where
$$n_{x,y} = \operatorname{card}\,\lbrace A \subseteq X \mid x\in A \mathrm{~and~} y\not\in A\rbrace.$$
Each of the sets we're taking the cardinality of contains exactly $1/4$ of the subsets of $X.$ In particular, each $m_{x,y}$ equals the corresponding $n_{x,y},$ and it follows that your two sums are equal.
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In fact, the two sums are both equal to $2^{n-2} \sum_{x\neq y} f(x)\; f(y),$ where $n$ is the cardinality of $X.$  (This is true since there are $2^n$ subsets of X.)
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By the way, this doesn't really have anything to do with the function $f.$  It would probably be better to formulate it either as a formal polynomial identity in $n$ variables $x_1, \dots x_n\mathrm{,}$ as follows:
\begin{align}\sum_{A \subseteq \lbrace 1, \dots, n\rbrace} \sum_{\substack{i,j \in A
\\\\i \neq j}} x_i\, x_j &=\!\!\!\!\!\sum_{A \subseteq \lbrace 1, \dots, n\rbrace} \; \sum_{i \in A,\,j \not \in A} x_i \, x_j
\\\\&=2^{n-1}\!\!\!\!\sum_{ 1\,\le\,i\,<\,j\,\le \,n} x_i \, x_j,
\end{align}
or, even better, as a formal polynomial identity in the $n^2-n$ variables $x_{i,\,j}$ with $1 \le i \le n, 1 \le j \le n, $ and $i\neq j\mathrm:$
\begin{align}\sum_{A \subseteq \lbrace 1, \dots, n\rbrace}\; \sum_{\substack{i,j \in A
\\\\i \neq j}} x_{i,\,j} &=\!\!\! \sum_{A \subseteq \lbrace 1, \dots, n\rbrace} \; \sum_{i \in A,\,j \not \in A} x_{i,\,j}
\\\\&=2^{n-2}\sum_{i\,\ne\,j} x_{i,\,j}.
\end{align}
These can be proven in the same way as the proof above.  Your identity comes from either of these by writing $X=\lbrace s_1,\dots s_n\rbrace,$ and setting each $x_i = f(s_i)$ in the first one, or setting $x_{i,j} = f(s_i)\;f(s_j)$ in the second one.
