There is also a quick abstract proof via representation theory: $S^2_0(\mathbb{R}^4)$ is a 9-dimensional representation of $\mathrm{SO}(4)/\{\pm I_4\}\simeq \mathrm{SO}(3)\times\mathrm{SO}(3)$ and, hence, must be a sum of tensor products of representations of the two simple factors. Each of the two (commuting) $\mathrm{SU}(2)$s in $\mathrm{SO}(4)$ actacts irreducibly on $\mathbb{R}^4$, so they each only preserve the one quadratic form, which is not in $S^2_0(\mathbb{R}^4)$. Hence each $\mathrm{SO}(3)\times\mathrm{SO}(3)$-irreducible piece of $S^2_0(\mathbb{R}^4)$ must be a sum of tensor products of nontrivial representations of the two factor $\mathrm{SO}(3)$s. Meanwhile, the lowest dimension nontrivial representation of $\mathrm{SO}(3)$ is $\mathbb{R}^3$. Hence $S^2_0(\mathbb{R}^4)$ must be isomorphic to the tensor product of the $3$-dimensional irreducible representations of the two factors.
A similar argument shows that $\Lambda^2(\mathbb{R}^4)$$\Lambda^2(\mathbb{R}^4)\simeq\mathbb{R}^6$, which is also a representation of $\mathrm{SO}(4)/\{\pm I_4\}\simeq \mathrm{SO}(3)\times\mathrm{SO}(3)$, must be a sum of two irreducible pieces $\Lambda^2_\pm\simeq\mathbb{R}^3$, each of which is the tensor product of the trivial representation of one factor $\mathrm{SO}(3)$ with the $3$-dimensional representation of the other factor $\mathrm{SO}(3)$.
Hence $S^2_0(\mathbb{R}^4)\simeq \Lambda^2_+\otimes\Lambda^2_-$ as $\mathrm{SO}(4)/\{\pm I_4\}$-modules.