Here is a revised partial answer and some comments:
It seems that you are asking for some kind of isomorphism between $S^2(\mathbb{F}^d)$ and $\Lambda^2(\mathbb{F}^{d+1})$ that would 'have the greatest symmetry', in the sense that it would commute with some group action on $\mathbb{F}^d$ and $\mathbb{F}^{d+1}$, the bigger the better, so that it would be as independent of a choice of basis as possible.
Ryan's suggestion of a basis-dependent identification has a kind of symmetry based on the symmetric group $S_d$, but it isn't induced by a corresponding symmetry of the underlying vector spaces $\mathbb{F}^d$ and $\mathbb{F}^{d+1}$ because, if you permute the bases of these spaces, you'll see that the $x_ix_j$ permute among themselves, but the $\mathrm{d}x_i\wedge\mathrm{d}x_j$ permute and change signs at the same time, so it's not really based on the geometry of the underlying vector spaces.
It seems that what you want to find a subgroup $H\subset\mathrm{GL}(d,\mathbb{F})\times \mathrm{GL}(d{+}1,\mathbb{F})$ such that $S^2(\mathbb{F}^d)$ and $\Lambda^2(\mathbb{F}^{d+1})$ would be isomorphic as $H$-modules, and you'd want $H$ to be maximal with this property. It is an interesting questions as to what the maximal such subgroups $H$ could look like.
Here is an example of the kind of thing that you might consider as an answer to your question:
For a $2$-dimensional vector space $V$, there is a natural isomorphism
$$
\Lambda^2\bigl(S^2(V)\bigr) = S^2(V)\otimes\Lambda^2(V),
$$
i.e., this isomorphism is $\mathrm{GL}(V)$-equivariant, and this provides an isomorphism between the skew-symmetric $2$-forms on $S^2(V^*)\simeq \mathbb{F}^3$ and the quadratic forms on $V^*\simeq \mathbb{F}^2$, twisted by the determinant $\Lambda^2(V)$, a $1$-dimensional vector space.
You can get rid of this twisting if you fix a volume form on $V$ and consider only volume perserving automorphisms, i.e., $\mathrm{SL}(V)$. Alternatively, you can consider $H = \mathrm{SL}(V)\times \mathbb{F}^\times$ and let the $\mathrm{SL}(V)$ factor act on $S^2(V)$ and $V$ as above, but let an invertible element $\lambda\in F^\times$ act on $V\times S^2(V)$ as $(\lambda,\lambda)$ (i.e., with equal weights, instead of as $(\lambda,\lambda^2)$), and this will give an $H$ acting on $V\times S^2(V)$ for which $S^2(V)$ and $\Lambda^2\bigl(S^2(V)\bigr)$ are isomorphic as $H$-modules on the nose (with no $\Lambda^2(V)$-twisting).
This generalizes to the case of all $d$, because, by the Clebsch-Gordan formulae,
there is an isomorphism of $\mathrm{GL}(V)$-modules
$$
S^2\bigl(S^{d-1}(V)\bigr)\otimes\Lambda^2(V) = \Lambda^2\bigl(S^{d}(V)\bigr).
$$
Since $S^{d-1}(V)\simeq\mathbb{F}^d$ and $S^{d}(V)\simeq\mathbb{F}^{d+1}$, this seems to correspond to the kind of thing you are thinking about. (As before, you can get rid of the $\Lambda^2(V)$-factor by restricting to $\mathrm{SL}(V)$ and instead making the scalars act with equal weights on $S^{d-1}(V)$ and $S^{d}(V)$, instead of with their 'natural' weights $d{-}1$ and $d$. The vector spaces $S^{d-1}(V)$ and $S^{d}(V)$ remain irreducible under this action.)
Note, however, that when $k\ge2$, the module $S^2\bigl(S^{k}(V)\bigr)$ is no longer an irreducible $\mathrm{SL}(V)$ module, instead, we have
$$
S^2\bigl(S^{k}(V)\bigr)\simeq \Lambda^2\bigl(S^{k+1}(V)\bigr)
\simeq \bigoplus_{0\le j\le k/2} S^{2k-4j}(V).
$$
A good test case would be $d=3$. I suspect that, in this case, the scalar-modified embedding as above of $H=\mathrm{GL}(V)$ into
$$
\mathrm{GL}\bigl(S^2(V)\bigr)\times \mathrm{GL}\bigl(S^3(V)\bigr)
\simeq \mathrm{GL}(3,\mathbb{F})\times \mathrm{GL}(4,\mathbb{F})
$$
makes $\mathrm{GL}(V)$ into a maximal subgroup of $\mathrm{GL}(3,\mathbb{F})\times \mathrm{GL}(4,\mathbb{F})$ for which $S^2(\mathbb{F}^3)$ and $\Lambda^2(\mathbb{F}^4)$ are isomorphic as $H$-modules, and I can show that any such $H$ that is a connected Lie group that contains a simple Lie subgroup must be this one, up to isomorphism. What happens in higher dimensions, I don't know.
