= Solution
The <dominance order on partitions> is defined by
$$
\lambda\unrhd\mu\quad\Longleftrightarrow\quad
\sum_{i=1}^r\lambda_i\ge\sum_{i=1}^r\mu_i\quad\text{for every }r.
$$
Write $\lambda\triangleright\mu$ when the shapes are distinct as well. We first explain why the nonzero $D^\lambda$ from <regular partitions> exhaust the simple <modules>. The argument also provides the required constraint on <composition factors>.
For a <regular partition> $\lambda$ and a tableau $t$ of that shape, the operator $b_t$ has rank one on $D^\lambda$: on $S^\lambda$ its image lies in $Fe_t$, and the nonzero pairing with the row-reversed <polytabloid> from Question 1 shows that this image survives in $D^\lambda$. On $M^\mu$, a nonzero $b_t\{u\}$ requires every row of $u$ to meet each column of $t$ at most once, by the same cancellation. Consequently the first $r$ rows of $u$ contain at most $\min(r,\lambda'_j)$ entries from column $j$. Summing gives
$$
\sum_{i=1}^r\mu_i\le\sum_j\min(r,\lambda'_j)
=\sum_{i=1}^r\lambda_i.
$$
Thus nonzero column action forces $\lambda\unrhd\mu$. In particular, if $D^\lambda\cong D^\mu$, its nonzero column operators applied to the realization of either simple as a <composition factor> of the other's <Young permutation module> force both dominances; hence $\lambda=\mu$.
Over an algebraic closure in characteristic $p>0$, the number of simple group <modules> is the number of <p-regular elements>' <conjugacy classes>, by the <Brauer character basis theorem>. Such classes in $S_n$ have cycle lengths not divisible by $p$. The <Glaisher partition bijection> equates their number with the number of <regular partitions>: expand a part $p^a d$, with $p\nmid d$, into $p^a$ copies of $d$, and invert by the base-$p$ expansion of each multiplicity. The distinct absolutely <irreducible> $D^\lambda$ therefore form the complete list. Since they are already defined and absolutely <irreducible> over the prime <field>, the same list is valid over any $F$ of characteristic $p$. In <characteristic zero>, the analogous count uses all <partitions of an integer> and all <conjugacy classes>.
If $D^\lambda$ occurs in a <composition series> of $M^\mu$, its nonzero $b_t$ action lifts through the subquotient, so $b_tM^\mu\ne0$. The inequality above proves $\lambda\unrhd\mu$. When $\lambda=\mu$ is regular, $b_t$ has rank one on $M^\mu$ and rank one on $D^\mu$. An operator preserving a finite filtration has rank at least the sum of its ranks on the successive quotients: relative to an adapted basis its matrix is block triangular, and nonzero pivot minors of the diagonal blocks combine to a nonzero minor. Therefore $D^\mu$ can occur at most once. It occurs at least once as the quotient of $S^\mu\subset M^\mu$. We have proved
$$
\boxed{[M^\mu:D^\lambda]\ne0\Rightarrow\lambda\unrhd\mu,
\qquad [M^\mu:D^\mu]=1\text{ if }\mu\text{ is regular}.}
$$
For a nonregular $\mu$, equality is impossible because $D^\mu=0$.
Over $\mathbb Q$, <Young's rule> gives the exact decomposition
$$
M^\mu\cong\bigoplus_{\lambda\vdash n}(S^\lambda)^{\oplus K_{\lambda\mu}}.
$$
The <Kostka number> $K_{\lambda\mu}$ counts <Semistandard Young tableaux> of shape $\lambda$ and content $\mu$: entry $i$ occurs $\mu_i$ times, rows weakly increase, and columns strictly increase. In particular $K_{\mu\mu}=1$ and nonzero multiplicity requires dominance.
For the requested induction, products mean induction of the <external tensor product of group representations> from $S_5\times S_2$ to $S_7$: on $V\boxtimes W$, the pair $(g,h)$ acts as $v\otimes w\mapsto gv\otimes hw$. One can compute using <Young's rule> alone. The two-row <permutation> decompositions give
$$
[S^{(3,2)}]=[M^{(3,2)}]-[M^{(4,1)}],
$$
where subtraction is in the <ordinary character> group. Reordering row blocks gives isomorphic <permutation modules>. Inducing after multiplication by the trivial $S^{(2)}$ gives $[M^{(3,2,2)}]-[M^{(4,2,1)}]$. For a direct <Kostka number> calculation, the differences $K_{\nu,(3,2,2)}-K_{\nu,(4,2,1)}$ are one for
$$
\nu=(5,2),(4,3),(4,2,1),(3,3,1),(3,2,2)
$$
and zero for every other <partition of an integer> of seven. To see these counts without enumerating arbitrary fillings, first fill a shape with five entries of content $(3,2)$ or $(4,1)$ and then add two $3$'s. The added cells must be a <horizontal strip>. Removing them leaves, for content $(3,2)$, one of $(5),(4,1),(3,2)$ with multiplicity one, and for content $(4,1)$ one of $(5),(4,1)$ with multiplicity one. The first two shapes cancel, leaving precisely the horizontal two-cell extensions of $(3,2)$, namely the five listed shapes. Consequently
$$
\boxed{\operatorname{Ind}_{S_5\times S_2}^{S_7}
(S^{(3,2)}\boxtimes S^{(2)})
\cong S^{(5,2)}\oplus S^{(4,3)}\oplus S^{(4,2,1)}\oplus S^{(3,3,1)}\oplus S^{(3,2,2)}.}
$$
The <hook-length formula> checks the <dimensions>: $14+14+35+21+21=105=\binom75\cdot5$.
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