= Solution
For the <degree-ten partition action of S6>, take unordered partitions of six letters into two triples. There are $\binom63/2=10$. Fix $A\mid A^c$. Its <stabilizer> is $(S_3\times S_3)\rtimes S_2=S_3\wr S_2$, permuting each triple and interchanging them. The <subgroup> generated by a 3-cycle in each triple has order nine, so is a <Sylow subgroup> of $S_6$. Any <normalizer> preserves its two three-point orbits; conversely all the indicated within-triple <permutations> and the swap normalize it. This proves the asserted <normalizer> exactly.
For any different partition $C\mid C^c$, $|A\cap C|$ is one or two after excluding the identical partition. Interchanging $C$ and $C^c$ identifies these two possibilities. The <stabilizer> of $A\mid A^c$ is transitive on each fixed intersection pattern, hence on all other nine partitions. \b[The degree-ten action is <two-transitive>.] It is also <faithful>: two distinct partitions have a common refinement with two singleton cells and two cells of size two. Every unordered pair of letters occurs as those singleton cells. A <permutation> fixing every partition therefore fixes every pair and consequently every letter.
For the <subgroup factorization from coset transitivity>, use <cosets> $Hg$ (right cosets) and right multiplication by $K$. The orbit of $H$ is precisely $\{Hk:k\in K\}$. This is every <coset> if and only if every $g$ can be written $hk$, proving
$$
\boxed{G=HK\quad\Longleftrightarrow\quad K\text{ is transitive on }H\backslash G.}
$$
A natural $S_5$ fixing one letter is transitive on the ten partitions: the triple containing the fixed letter is determined by any two of the other five. Taking $H=S_3\wr S_2$, $K=S_5$ gives $S_6=HK$; inversion gives the requested order as well,
$$
\boxed{S_6=S_5(S_3\wr S_2).}
$$
Now let $B$ be the binary <nondegenerate> <alternating bilinear form>. Its <binary quadratic refinements> form an <affine space>: the difference of two such forms is linear, so for a fixed $Q_0$ every one is uniquely
$$
Q_v(x)=Q_0(x)+B(v,x),\qquad v\in V.
$$
In a <symplectic basis> each refinement is
$$
Q(x,y)=\sum_{i=1}^m\bigl(x_iy_i+a_ix_i+b_iy_i\bigr),\qquad
\operatorname{Arf}(Q)=\sum_{i=1}^ma_ib_i\in\mathbb F_2.
$$
A two-dimensional summand with $(a_i,b_i)\ne(1,1)$ admits a <basis> of singular vectors paired to one and is hyperbolic; the $(1,1)$ summand is anisotropic. Two planes spanned by <anisotropic vectors for a form> together become two <hyperbolic planes of a quadratic form>: with anisotropic <basis> pairs $(e_1,f_1),(e_2,f_2)$, the singular vectors $e_1+e_2,f_1+e_2$ pair to one; their <orthogonal complement for a sesquilinear form> has singular paired <basis> $e_1+f_1+f_2,e_1+f_1+e_2+f_2$. Consequently every refinement is equivalent to either $m$ <hyperbolic planes of a quadratic form> or $(m-1)$ <hyperbolic planes of a quadratic form> and one anisotropic plane (all three nonzero vectors are <anisotropic vectors for a form>).
These two types are different. Indeed the character sum of each plane is $+2$ or $-2$, so
$$
\sum_{x\in V}(-1)^{Q(x)}=(-1)^{\operatorname{Arf}(Q)}2^m.
$$
Thus there are exactly two symplectic orbits $Q^+(B),Q^-(B)$, the signs denoting $\operatorname{Arf}=0,1$. Counting the three coefficient pairs with product zero and the one with product one gives
$$
\boxed{|Q^\pm(B)|=\frac{4^m\pm2^m}{2}=2^{m-1}(2^m\pm1).}
$$
The same character sum gives the corresponding numbers of singular vectors, including zero, in a fixed form of either type.
Fix a form $Q$ of type $\epsilon$. Completing the square by $x\mapsto x+v$ shows
$$
\sum_x(-1)^{Q_v(x)}=(-1)^{Q(v)}\sum_x(-1)^{Q(x)},\qquad
\epsilon(Q_v)=\epsilon(Q)(-1)^{Q(v)}.
$$
The <stabilizer> of $Q$ is exactly $O(V,Q)$. It sends the parameter $v$ by its natural linear action or inverse action according to the chosen left/right convention. Other forms of the same type correspond to nonzero singular vectors $v$, and all forms of opposite type correspond to nonsingular vectors $v$. For two nonzero vectors of the same $Q$ value, the map between their spans is an <isometry of a space with a form>. <Witt's lemma> extends it, so the orthogonal <stabilizer> is transitive on each of these two vector sets. This proves the <two-transitive binary quadratic-form actions>: the <symplectic group over a finite field> is transitive on each type and its point <stabilizer> is transitive on all other members of that type.
In particular $O^-_{2m}(2)$ is transitive on the plus orbit. Identify that orbit with the <cosets> (right cosets) of a plus <stabilizer>, using $Q^g(x)=Q(gx)$ as a right action. The <coset> criterion then proves
$$
\boxed{Sp_{2m}(2)=O^+_{2m}(2)O^-_{2m}(2).}
$$
Both genuine <two-transitive> actions require $m\ge2$. At $m=1$ the plus orbit has three forms and the minus orbit is a singleton; transitivity and the factorization remain true, while <two-transitive> action on the singleton is only vacuous if one's convention permits degree one. No special simplicity assertion is needed for this argument. At $m=2$ the orbit sizes are ten and six, linking these actions with $Sp_4(2)\cong S_6$ and the partition action above.
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