= Solution
For the standard tableau with second row $2,4$, the first entries in the top row are $1,3$, so its nontrivial columns are $(1,2)$ and $(3,4)$. Hence
$$
e_t=w_{24}-w_{14}-w_{23}+w_{13}.
$$
In $\phi_2(e_t)$ the coefficient of $x_{24n}$ is one: among those four pairs only $\{2,4\}$ is contained in $\{2,4,n\}$, since $n\geq6$. Thus the image is nonzero. Its restriction to the characteristic-zero <simple module> $S^{(n-2,2)}$ has zero kernel, giving the requested isomorphic submodule.
For $S^{(n-1,1)}$, choose the standard tableau with lower entry two and top-left entry one. Its <polytabloid> is $v_2-v_1$. Each three-subset containing $a$ occurs twice in $\phi_2\phi_1(v_a)$, once for each choice of its other two entries as the intermediate $b$. Consequently the coefficient of $x_{23n}$ in $\phi_2\phi_1(v_2-v_1)$ is two. This is nonzero in characteristic zero. Simplicity again makes the restriction injective, giving \b[a copy of $S^{(n-1,1)}$].
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