To describe an invariant trivector in dimension 8 geometrically

Here's another very nice (but still algebraic) interpretation that explains some of the geometry: Recall that $\operatorname{SL}(2,\mathbb{C})$ has a $2$-to-$1$ representation into $\operatorname{SL}(3,\mathbb{C})$ so that the Lie algebra splits as $$ {\frak{sl}}(3,\mathbb{C}) = {\frak{sl}}(2,\mathbb{C})\oplus {\frak{m}} $$ where ${\frak{m}}$ is the ($5$-dimensional) orthogonal complement of ${\frak{sl}}(2,\mathbb{C})$ using the Killing form of ${\frak{sl}}(3,\mathbb{C})$. Note that ${\frak{m}}$ is an irreducible ${\frak{sl}}(2,\mathbb{C})$-module, and that every element $x\in {\frak{sl}}(3,\mathbb{C})$ can be written uniquely as $x = x_0 + x_1$ with $x_0\in {\frak{sl}}(2,\mathbb{C})$ and $x_1\in{\frak{m}}$. Note also that $[{\frak{m}},{\frak{m}}]= {\frak{sl}}(2,\mathbb{C})$.

This defines the desired pairing ${\frak{sl}}(2,\mathbb{C})\times \bigwedge\nolimits^2({\frak{m}})\to\mathbb{C}$: Send $(x_0,y_1,z_1)$ to $\operatorname{tr}(x_0[y_1,z_1])$. Of course, this makes the $\operatorname{SL}(2,\mathbb{C})$-invariance of the pairing obvious.

For a purely geometric construction, see further below, after the following algebraic considerations.

There is a Wronskian isomorphism which as a particular case says that the second exterior power of $R_4$ is isometric to the second symmetric power of $R_3$. So the invariant in question is $I(Q,C)$, a joint invariant in a binary quadratic $Q$ and a binary cubic $C$, which is linear in $Q$ and quadratic in $C$. This is indeed unique up to scale and is given in classical symbolic notation (see, e.g., Grace and Young) by $$ (ab)(ac)(bc)^2 $$ where $Q=a_{x}^{2}$ and $C=b_{x}^{3}=c_{x}^{3}$.

Another construction is to start from the binary discriminant, and polarize it to get a bilinear form (the unique invariant one on $R_2$), and apply this bilinear form to $Q$ and the Hessian of $C$.

If one does not want to use the Wronskian isomorphism then the invariant would be $J(Q,F_1,F_2)$, trilinear in the quadratic $Q$ and the two binary quartics $F_1,F_2$. It would satisy the antisymmetry $J(Q,F_2,F_1)=-J(Q,F_1,F_2)$ and would be given in symbolic form by $$ (ab)(ac)(bc)^3 $$ where now $Q=a_{x}^{2}$, $F_1=b_{x}^{4}$, and $F_2=c_{x}^{4}$.

---

Geometric construction:

Consider $\mathbb{P}^1$ embedded by Veronese as a conic $\mathscr{C}$ in $\mathbb{P}^2$. A binary quadratic $Q$ corresponds to a point in $\mathbb{P}^2$. A binary cubic $C$ corresponds to a divisor or an unordered collection of three points $\{P_1,P_2,P_3\}$ on $\mathscr{C}$. Let $T_1, T_2, T_3$ be the tangents to the conic at $P_1,P_2,P_3$. Consider the points of intersection $T_1\cap P_2P_3$, $T_2\cap P_1P_3$, $T_3\cap P_1P_2$. They are aligned and thus define a line $L$. The vanishing of the invariant $I(Q,C)$ detects the situation where the point $Q$ is on the line $L$. I don't remember if the collinearity result I mentioned has a name, but it is a degenerate case of Pascal's Theorem.