Given a real Hilbert space $\mathcal{H}$.
Consider symmetric forms: $$s:\mathcal{H}\times\mathcal{H}\to\mathbb{R}:\quad s(\psi,\varphi)=s(\varphi,\psi)$$
By polarization one obtains: $$s(\varphi,\psi)=\frac{1}{4}\{s(\varphi+\psi,\varphi+\psi)-s(\varphi-\psi,\varphi-\psi)\}$$
Consider arbitrary forms: $$s:\mathcal{H}\times\mathcal{H}\to\mathbb{R}:\quad s(\psi,\varphi)\neq s(\varphi,\psi)$$
Can one reconstruct them?
Let $X$ be a real vector space. Let $\mathcal{B}$ be the real vector space of bilinear forms $X \times X \to \mathbb{R}$ with the natural pointwise operations.
Now $\mathcal{B}$ is the internal direct sum of the subspaces $\mathcal{B}_s$ and $\mathcal{B}_a$ of symmetric forms and antisymmetric forms. A bilinear form $f$ is symmetric if $f(x,y) = f(y,x)$ for all $x,y \in X$. A bilinear form is antisymmetric if $f(x,y) = -f(y,x)$ for all $x,y \in X$. Equivalently, a bilinear form is antisymmetric iff $f(x,x) = 0$ for all $x \in X$.
Given any bilinear form $f \in \mathcal{B}$, the symmetric and antisymmetric parts $f_s \in \mathcal{B}_s,f_a \in \mathcal{B}_a$ such that $f = f_s + f_a$ are given by \begin{align*} f_s(x,y) = \frac{f(x,y) + f(y,x)}{2} && f_a(x,y) = \frac{f(x,y) - f(y,x)}{2} \end{align*}
Associated to each bilinear form $f \in \mathcal{B}$ is the quadratic form $Q_f : X \to \mathbb{R}$ given by $Q_f(x) = f(x,x)$. Note the assignment $f \mapsto Q_f$ is linear i.e. $Q_{af + bg} = a Q_f + b Q_g$ for all bilinear forms $f,g$ and all scalars $a,b$.
I think what you are asking is "to what extent can I recover $f$ from $Q_f$?" You already know that the restriction of $f \mapsto Q_f$ to $\mathcal{B}_s$ is invertible, with inverse given by polar decomposition. But, note as well that $Q_f = 0$ if and only if $f$ is antisymmetric. So, the kernel of $f \mapsto Q_f$ exactly equals $\mathcal{B}_a$.
So, if $f = f_s + f_a$ is some bilinear form, then $Q_f = Q_{f_s}$. That is, you can recover the symmetric part of $f$ from $Q_f$, but you can never learn anything about the antisymmetric part.