Maybe this is something like what you have in mind:

Let $V$ be a (complex) vector space of dimension $3$. It is easy to show that a generic subspace $P\subset S^2(V^*)$ of dimension $4$ can be written as
$$
P = \mathrm{span}\{\ {x_1}^2,\ {x_2}^2,\ {x_3}^2,\ x_1x_2{+}x_2x_3{+}x_3x_1\ \}
$$
for some basis $x_1,x_2,x_3$ of $V^*$. (There are only a finite number of other 'nongeneric' cases to be handled.)

It seems that you are asking for a 'natural' basis for the kernel of the multiplication map
$$
\mu:P\otimes S^2(V^*)\longrightarrow S^4(V^*)
$$
Now, $P\subset S^2(V^*)$ is invariant under permutations of the basis $x_1,x_2,x_3$, a group isomorphic to $S_3$. The action of $S_3$ on $S^2(V^*)$ then has a natural complement to $P$, which is the space
$$
W = \mathrm{span}\{\ \ x_1x_2{-}x_2x_3,\ x_2x_3{-}x_3x_1\ \}.
$$
As representations of $S_3$, $W$ is irreducible and $P \simeq \mathbb{C}\oplus\mathbb{C}\oplus W$.

Now, we have the natural decomposition
$$
P\otimes S^2(V^*) = P\otimes (P \oplus W)
= \Lambda^2(P)\oplus S^2(P)\oplus P\otimes W.
$$
Clearly, $\mu\bigl(\Lambda^2(P)\bigr) = 0$ (these are the relations you are calling the 'Koszul relations'), so we need to examine the other two pieces. It is easy to check that $\mu$ is surjective, so, indeed, there is a $3$-dimensional kernel of the map
$$
\mu: S^2(P)\oplus P\otimes W\to S^4(V^*).
$$
It's obvious that $\mu$ is injective on $S^2(P)$ and it's easy to check that it's injective on $P\otimes W$. The two $\mu$-images intersect in a space of dimension $3$ that is invariant under $S_3$, and it is easy to see from this that the kernel $K$ splits as an $S_3$-module as a sum $K\simeq \mathbb{C}\oplus W$. Hence, there is one relation that is invariant under the action of $S_3$ and the complement is an irreducible $S_3$-module of dimension $2$. You can write them out without difficulty, but they all involve a significant number of terms.