If, as you say, you have complete freedom to pick an $F$ yourself, then I think that it is possible with a large enough vector space $V$.  I'll construct $F$ and $V$ together, and then construct the bilinear pairing last.  Let $V_0 = \mathbb{R}$ with its basis vector $1$.  Then let $V_{n+1}$ be the direct sum of $V_n$ and the vector space $W_n$ of formal linear combinations of elements of $V_n \setminus V_{n-1}$, where in this formula $V_{-1} = \emptyset$.  If $v \in V_n \setminus V_{n-1}$, let $[v]$ denote the corresponding element in $W_n \subset V_{n+1}$.  Let $V$ be the union of all $V_n$, and let $F(v) = [v]$.  Note that every $v \in V$ has a degree $d(v)$, by definition the first $n$ such that $v \in V_n$.

To construct the pairing, let $\langle 1,1 \rangle = 1$.  We need to choose values of $\langle e,f \rangle$ for every other unordered pair of basis vectors $e,f$.  I claim that your constraints are triangular with respect to degree, in other words that the values can be constructed by induction.  Also the diagonal values $\langle e, e \rangle$ are unrestricted.  To see this, consider your equation
$$\langle F(x), F(x) \rangle + \langle F(y),F(y) \rangle - 2\langle F(x), F(y) \rangle = \langle F(x) - F(y), F(x) - F(y) \rangle = f(\langle x-y, x-y \rangle)$$
with $x \ne y$.  By construction, the arguments of the cross-term $\langle F(x), F(y) \rangle$ are both basis vectors, and only occur once for any given $x$ and $y$. Let's say that $\max(d(x), d(y)) = n$.  Then $d(x-y) \le n$.  In defining the inner product on $V_{n+1}$, the right side of your equation is already chosen, two terms on the left are unrestricted, and the third term can be chosen to satisfy the equality.

You could economize in the construction a little bit by setting $V_{-1} = \{0\}$ instead.  In this case, the diagonal terms $\langle F(x), F(x) \rangle$ are not unrestricted.  Instead, you should set
$$\langle F(x), F(x) \rangle  = f(\langle x, x \rangle),$$
taking $y = 0$ in your equation.