There's a 'quasi-normal form' on a dense open set that can be described without too much trouble. Here's one way to do it.

First, recognize that we are looking for a normal form for elements $Q\in W=\bigl[S^2(\Lambda^2(V^\ast))\otimes V^\ast \bigr]_0$ under the action of $G=\mathrm{GL}(V)$, where $V$ is a vector space of dimension $3$ and $W$ is the kernel of the $G$-module mapping $$
S^2(\Lambda^2(V^\ast))\otimes V^\ast\hookrightarrow
\Lambda^2(V^\ast)\otimes \Lambda^2(V^\ast)\otimes V^\ast\to \Lambda^2(V^\ast)\otimes\Lambda^3(V^\ast)$$
defined to be the natural inclusion followed by fully skewsymmetrizing in the last two factors. This $W$ is an irreducible $G$-module of dimension $15$.

Now, it makes things a little easier to follow if one recognizes that there is a canonical isomorphism $\Lambda^2(V^\ast) = V\otimes \Lambda^3(V^\ast)$ (given by the obvious contraction), so that $W$ can also be understood as a subspace of $S^2(V)\otimes V^\ast\otimes \bigl(\Lambda^3(V^\ast)\bigr)^2$. In fact $W = W_0\otimes \bigl(\Lambda^3(V^\ast)\bigr)^2$, where $W_0\subset S^2(V)\otimes V^\ast$ is the $15$-dimensional subspace that is the kernel of the trace mapping $S^2(V)\otimes V^\ast\to V$. So I am going to think of $Q$ as an element of $W_0\otimes \bigl(\Lambda^3(V^\ast)\bigr)^2$.

The vector space $U = V^\ast\otimes \bigl(\Lambda^3(V^\ast)\bigr)^2$ has dimension $3$, and by pairing $Q$ with elements in $U^\ast$, one can generate a subspace $\delta Q\subset S^2(V)$ of dimension at most $3$; say that $Q$ is of *full quadratic rank* if $\delta Q$ has dimension $3$, i.e., $\delta Q$ lies in $\mathrm{Gr}_3\bigl(S^2(V)\bigr)$. The set of $3$-dimensional subspaces of a $6$-dimensional vector space has dimension $9$, and there is another way to generate a $9$-parameter family $3$-dimensional subspaces of $S^2(V)$, namely, if $C\in S^3(V)$ is a nondegenerate cubic, thought of as a cubic polynomial function on $V^\ast$, then we can let $\partial C\subset S^2(V)$ denote the $3$-dimensional space of its partial derivatives. (In fact, 'nondegenerate' in this context means exactly that $\partial C$ has dimension $3$, the maximum dimension possible.)

Now, these two methods of generating a $3$-dimensional subspace of $S^2(V)$ are related: For *any* $3$-dimensional subspace $P\subset S^2(V)$, there is always a cubic $C$ such that $\partial C\subset P$, and, most of the time, this cubic is unique up to multiples and satisfies $\partial C = P$. One can see this as follows: There is a natural exact sequence
$$
0\to S^3(V)\to S^2(V)\otimes V\to V\otimes \Lambda^2(V)\to \Lambda^3(V)\to 0.
$$
In particular, the image of the map $S^2(V)\otimes V\to V\otimes \Lambda^2(V)$ has dimension $8$. If $P\subset S^2(V)$ has dimension $3$, then the restriction of this map to $P\otimes V\to V\otimes \Lambda^2(V)$ has rank at most $8$, so there must be at least a $1$-dimensional kernel, i.e., there must be a nonzero $C\in S^3(V)$ whose image under the above mapping lies in $P\otimes V$. Say that a $3$-dimensional subspace $P\subset S^2(V)$ is *uniquely partial* if the kernel of $P\otimes V\to V\otimes \Lambda^2(V)$ has dimension $1$. The set of uniquely partial $P$ is a dense open set in $\mathrm{Gr}_3(S^2(V))$. In particular, for a dense open set of $Q\in W$, the subspace $\delta Q$ will be of dimension $3$ and uniquely partial. Let us say that $Q$ is *nonsingular* if, in addition, the cubic $C\in S^2(V)$ (unique up to multiples) such that $\partial C = \delta Q$ is nonsingular. The set of nonsingular $Q$ is a dense open set in $W$.

Finally, put the nonsingular $C$ associated (uniquely up to multiples) to a given nonsingular $Q$ in normal form, i.e., take a basis $e_i\in V$ such that
$$
C = {e_1}^3+{e_2}^3+{e_3}^3 + 6\sigma\ e_1e_2e_3\ ,
$$
where $\sigma\not=-\tfrac12$ is a real number. (In fact, one needs to disallow a few more values of $\sigma$ in order to make sure that $\partial C$ is uniquely partial, but I'll leave that to the reader. For example, $\sigma=0$ is not allowed.) The basis $e_i$ is uniquely determined by $C$ up to permutations. (Of course, $C$ is only determined up to a multiple, so the basis $e_i$ is only defined by $Q$ up to permutation and simultaneous scaling by a real number. This $S_3\times \mathbb{R}^*$-ambiguity is the reason that the normal form is only 'quasi-normal', as we will see.)

Let $e^i$ be the dual basis. Now, by definition, $Q$ lies in $\partial C\otimes V^\ast\otimes S^2(\Lambda^3(V^\ast))$, so there are numbers such that
$$
Q = \tfrac13 b_{ij} \frac{\partial C}{\partial e_i}\otimes e^j
\otimes (e^1\wedge e^2\wedge e^3)^{\otimes 2}
$$
The $9$ entries of $b = (b_{ij})$ are subject to three linear equations caused by the trace relation (i.e., the second Bianchi identity), and these relations are found to be $b_{ii}+\sigma(b_{jk}+b_{kj})=0$, where $(i,j,k)$ is any even permutation of $(1,2,3)$. They must also satisfy $\det b\not=0$, since, otherwise, $Q$ would be degenerate.

Obviously, these relations are invariant under the symmetric group and scaling. One could normalize the scaling away (up to a $\pm1$) by requiring that the sum of the squares of the $b_{ij}$ be equal to $1$, and this would leave a finite group $S_3\times \lbrace\pm1\rbrace$
that preserves this quasi-normal form. This brings the number of free parameters down to $6$, namely $\sigma$ and the $5$-sphere of the normalized $b_{ij}$ (which lie in a canonical $6$-dimensional subspace that depends on $\sigma$). One could then use the finite group to normalize things further or use the invariant theory of this finite group to find invariant combinations of these quantities that will yield invariants of the original $\mathrm{GL}(V)$-action on the dense open set consisting of the nonsingular $Q$s.

With a little more work, one could actually get quasi-normal forms for all of the orbits in $W$, but that can get complicated.