The span of the monomials of the form $x_i^2x_jx_k$ is the Young permutation module $M^{(2,1,1)}$. (<em>Proof.</em> Observe that $x_1^2x_2x_3$ has stabiliser $\langle (2,3)\rangle$, so the relevant Young subgroup is $S_2 \times S_1 \times S_1$.) Using Kostka numbers (equivalently, multiplicities of Schur functions in complete symmetric funtions) this decomposes as

$$M^{(2,1,1)} \cong S^{(2,1,1)} \oplus S^{(2,2)} \oplus S^{(3,1)} \oplus S^{(3,1)} \oplus S^{(4)}.$$

(I'm using superscripts for Specht's modules since this is the notation I'm used to.) 

To make this explicit, Specht's original construction of Specht modules shows that $S^{(2,1,1)}$ is generated by the product of Vandermonde determinants

$$\left| \begin{matrix} 1 & 1 & 1 \\ x_1 & x_2 & x_3 \\ x_1^2 & x_2^2 & x_3^2 \end{matrix} \right| x_1 $$

and it's clear that the unique trivial submodule is spanned by $x_1^2x_2x_3 + x_1^2x_2x_4 + \cdots + x_2x_3x_4^2$, i.e. the sum of all monomials whose exponents are $2$, $1$, $1$, $0$ 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 must be injective. I'll give some details here. By this theory,

$$\mathrm{Hom}_{\mathbb{C}S_4}(S^{(3,1)}, M^{(2,1,1)})$$

is spanned by the semistandard homomorphisms for the two semistandard tableaux of shape $(3,1)$ and content $(2,1,1)$: these have rows $112,3$ and $113$, $2$, respectively. Each such homomorphism extends to $M^{(3,1)}$. Taking as a model for $M^{(3,1)}$ the natural permutation module $\langle e_1,e_2,e_3,e_4\rangle$, and using tableaux of shape $(3,1)$ with content $(2,1,1)$ as a model for $M^{(2,1,1)}$ (as in the semistandard setup), the homomorphisms are defined by (homomorphic extension of) $e_4 \mapsto 112,3 + 121,3 + 211,3$ and $e_4 \mapsto 113,2 + 131,2 + 311,2$. (Here the right-hand side summands are tableaux of shape $(3,1)$.) In the polynomial model, these becomes

$$e_4 \mapsto x_1x_2x_3^2 + x_1x_3x_2^2 + x_2x_3x_1^2$$

and

$$e_4 \mapsto (x_1x_2+x_2x_3+x_3x_1)x_4^2,$$

respectively. For $(2,2)$, there is a unique semistandard tableaux of shape $(2,2)$ and content $(2,1,1)$, namely $11,23$ and the corresponding homomorphism defined on a generator for 

$$M^{(2,2)} \cong \langle \{1,2\}, \ldots, \{3,4\} \rangle$$ 

is $\{1,2\} \mapsto x_1x_2(x_3^2+x_4^2)$. One then has to restrict these homomorphisms to $S^{(3,1)} \subseteq M^{(3,1)}$ and $S^{(2,2)} \subseteq M^{(2,2)}$ to get submodules of $M^{(2,1,1)}$ as in the claimed decomposition.