I think that it is possible with a large enough vector space $V$. I first misread the question, and constructed something where the inner product depends on $f$ while the mapping $F$ does not. The construction can be adapted to the true question as stated, so I'll still give it first as a warmup.

### Version 1

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 $x \in V_n \setminus V_{n-1}$, let $[x]$ denote the corresponding element in $W_n \subset V_{n+1}$. Let $V$ be the union of all $V_n$, and let $F(x) = [x]$. Note that every $x \in V$ has a degree $d(x)$, by definition the first $n$ such that $x \in V_n$.

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.

### Version 2

Suppose instead that the inner product is to be fixed and instead $F$ can change with $f$. In this case, let $W_n$ be the vector space of formal linear combinations of elements of $(V_n \setminus V_{n-1}) \times \mathbb{R}^\mathbb{R}$, and as before let $V_{n+1} = V_n \oplus W_n$. In this case, $W_n$ has a basis vector $[x,f]$ for every $f$ and every suitable $v$. For any fixed $f$, define $F(x) = [x,f]$.

As before, say that $\langle 1,1 \rangle = 1$ and that $\langle [x,f], [x,f] \rangle$ is unrestricted. Also $\langle [x,f], [y,g] \rangle$ is unrestricted when $f \ne g$, for all $x$ and $y$. Finally, as before,
$$\langle F(x), F(y) \rangle = \langle [x,f], [y,f] \rangle$$
with $x \ne y$ is uniquely determined by induction on $\max(d(x),d(y))$.

### Version 3

Ady reminds me that the second version still misses the condition that the bilinear form on $V$ should be non-degenerate. I think that the same trick works a third time: We can just enlarge $V$ to also guarantee this condition. This time let $W_n$ be as in the second version, and let
$$V_{n+1} = V_n \oplus W_n \oplus V_n^*,$$
where $V_n^*$ is the (algebraic) dual vector space to $V_n$.
Define the bilinear form on $V_n \oplus W_n$ as in version 2, and define $F$ as in version 2. The bilinear form on $V_n^*$ is unrestricted, and so is the bilinear pairing between $V_n^*$ and $W_n$. Finally the bilinear pairing between $V_n^*$ and $V_n$ should be the canonical pairing $\langle \phi, x \rangle = \phi(x)$. This guarantees that for every vector $x \in V_n$, there exists $y \in V_{n+1}$ such that $\langle y,x \rangle = 1$.

Every version of the construction is cheap in the sense that the image of $F$ is a linearly independent set. Moreover, in the second and third versions, the image of $F$ is far from a basis. My feeling is that it is difficult to ask for much better than that in a universal construction.