The span of the monomials of the form $x_i^2x_jx_k$ is the Young permutation module $M^{(n-3,2,1)}$. (Proof. Observe that $x_1^2x_2x_3$ has stabiliser $\langle (2,3)\rangle \times S_{\{4,\ldots,n\}}$, so the relevant Young subgroup is $S_{n-3} \times S_2 \times S_1$.) Using Kostka numbers (equivalently, multiplicities of Schur functions in complete symmetric funtions) this decomposes as
$$M^{(n-3,2,1)} \cong S^{(n-3,2,1)} \oplus S^{(n-3,3)} \oplus S^{(n-2,1,1)} \oplus 2S^{(n-2,2)} \oplus 2S^{(n-1,1)} \oplus S^{(n)}$$
provided that $n \ge 6$. (I'm using superscripts for Specht modules since this is the notation I'm used to.)
To make this explicit, Specht's original construction of Specht modules shows that $S^{(n-3,2,1)}$ is generated by the product of Vandermonde determinants
$$\left| \begin{matrix} 1 & 1 & 1 \\ x_1 & x_2 & x_3 \\ x_1^2 & x_2^2 & x_3^2 \end{matrix} \right| \left| \begin{matrix} 1 & 1 \\ x_4 & x_5 \end{matrix}\right| $$
and it's clear that the unique trivial submodule is spanned by
$$x_1^2x_2x_3 + x_1^2x_2x_4 + \cdots + x_{n-2}x_{n-1}x_n^2,$$
i.e. the sum of all monomials whose exponents are $2$, $1$, $1$, $0, \ldots, 0$ in some order.
For the other factors it is harder to make them explicit, but this can be done by using semistandard homomorphisms (see for instance James' lecture notes). Since we're working in characteristic zero, any non-zero homomorphism from a Specht module must be injective. I'll give some details here. By this theory,
$$\mathrm{Hom}_{\mathbb{C}S_n}(S^{(n-1,1)}, M^{(n-3,2,1)})$$
is spanned by the semistandard homomorphisms for the two semistandard tableaux of shape $(n-1,1)$ and content $(n-3,2,1)$: these have rows $1\ldots122,3$ and $1\ldots123$, $2$, respectively. Each such homomorphism extends to $M^{(n-1,1)}$. Taking as a model for $M^{(n-1,1)}$ the natural permutation module $\langle e_1,\ldots, e_n\rangle$, the corresponding extended homomorphisms are
$$e_n \mapsto (x_1x_2^2 + x_1x_3^2 + \cdots + x_{n-2}x_{n-1}^2)x_n$$
and
$$e_n \mapsto (x_1x_2+x_1x_3+\cdots + x_{n-2}x_{n-1})x_n^2,$$
respectively. One then has to restrict these homomorphisms to $S^{(n-1,1)} \subseteq M^{(n-1,1)}$ (for instance it is generated by $e_n-e_1$) to get two submodules of $M^{(n-3,2,1)}$ isomorphic to $S^{(n-1,1)}$, as in the claimed decomposition.
