= Solution
For a <Young diagram> cell $(i,j)$, its <hook length> is one plus the number of cells to its right and below. The <hook-length formula> is
$$
f^\lambda=\frac{n!}{\prod_{(i,j)\in\lambda}h_{ij}}.
$$
Applying it to the two shapes gives
$$
\boxed{f^{(n-2,1,1)}=\frac{(n-1)(n-2)}2,\qquad
f^{(n-2,2)}=\frac{n(n-3)}2.}
$$
The shapes here require $n\ge4$.
A hook <tabloid> remembers its first-row set and the ordered singleton entries below it. The proposed map keeps that first-row set and merges the remaining entries into one unordered row. Thus it is well-defined on tabloids. Permuting entries before merging gives the same <tabloid> as merging before permuting, so it is a <module> homomorphism. Every target <tabloid> is reached by putting its second-row entries, in any order, into the singleton rows. Therefore it is surjective.
For $r=2$, the strictly dominating partitions of $\lambda=(n-2,1,1)$ are exactly $(n),(n-1,1),(n-2,2)$. Indeed dominance forces the first part to be at least $n-2$, and the possibilities for the remaining sum give precisely those shapes or $\lambda$ itself.
In <characteristic zero>, <Maschke's theorem> makes the <permutation module> semisimple, and <Young's rule> gives one copy of $S^\lambda$ and only strictly dominating additional shapes. The trivial multiplicity is one, since a transitive <permutation module> has a one-dimensional invariant space. For the standard <representation> $S^{(n-1,1)}$, <Frobenius reciprocity> identifies its multiplicity with the <dimension> of fixed vectors under the row stabilizer $S_{n-2}$. In its realization $\{(z_1,\ldots,z_n):\sum z_i=0\}$, those fixed vectors have the form $(c,\ldots,c,a,b)$ with $(n-2)c+a+b=0$, a two-dimensional space. Thus its multiplicity is two.
Finally $\dim M^\lambda=n(n-1)$. Subtracting the dimensions of one $S^\lambda$, one trivial <module> and two standard modules leaves $n(n-3)/2$, exactly one $S^{(n-2,2)}$. Therefore
$$
\boxed{M^{(n-2,1,1)}\cong
S^{(n-2,1,1)}\oplus S^{(n)}\oplus2S^{(n-1,1)}\oplus S^{(n-2,2)}.}
$$
The additional multiplicities are $1,2,1$ respectively. If the printed weak summation index is read literally alongside its separate $S^\lambda$, then $m_{\lambda,\lambda}=0$; the total multiplicity of $S^\lambda$ is one.
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