Because $H$ is the point stabilizer of $G$, the action of $G$ on $A$ corresponds to the action of $G$ on the right cosets of $H$ in $G$.
First, notice that if you know what each element of $G$ does to $a$, then you know what each element of $G$ does to any element of $A$: given $b\in A$, and $g\in G$, since the action is transitive you can find $x\in G$ with $a\cdot x = b$. Then $b\cdot g = (a\cdot x)\cdot g = a\cdot(xg)$, and since you know what every element does to $a$, you know what $a\cdot(xg)$ is.
Second: notice that two elements do the same thing to $a$ if and only if they are in the same right coset of $H$ in $G$: because $a\cdot g= a\cdot g'$ if and only if $(a\cdot g)\cdot g'^{-1} = (a\cdot g')\cdot g'^{-1}$, if and only if $a\cdot(gg'^{-1}) = a\cdot 1 = a$, if and only if $gg'^{-1}\in H$, if and only if $Hg = Hg'$.
So in fact, the way $G$ acts on $A$ is exactly the same as the way that $G$ acts on the right cosets of $H$ in $G$. This is what the first paragraph means when it says that "The points $A$ are in bijection with the (right) cosets of $H$ in $G$." To each point $b\in A$ you can associate the coset $Hg$, where $g$ is any element of $G$ such that $a\cdot g = b$. And if we use that association, then given $b\in A$ and $x\in G$, if $b\leftrightarrow Hg$ then $b\cdot x = (a\cdot g)\cdot x = a\cdot(gx)$, so if $b\leftrightarrow Hg$ then $b\cdot x \leftrightarrow Hgx$.
The cosets of $K$ in $G$ are unions of $H$-cosets, namely: since $K = \cup_{k\in K} Hk$, then $Kg = \cup_{k\in K} Hkg$.
Now identify the sets $Kg = \{ Hkg\mid k\in K\}$ with subsets of $A$. Since $H\neq K$ and $K\neq G$, each of these, as a subset of $A$, is a proper subset of $A$ that is not a singleton. Since distinct cosets of $K$ are disjoint, the sets $Kg$ are pairwise disjoint. Morevoer, if $x\in G$, then
$$(Kg)\cdot x = \{Hkg\mid k\in K\}\cdot x = \{Hkgx\mid k\in K\} = K(gx)$$
so $(Kg)\cdot x = Kg$ or $(Kg\cdot x)\cap Kg =\emptyset$.
That means that the cosets of $K$ are nontrivial blocks of $A$ under the action of $G$, so this shows that the action is imprimitive.
Hint: Notice that the cycle $(k,l)\in S_n$ for any $k,l\in \{1,2,..,n\}$.
Best Answer
Your conclusion is not correct. You can check that if $H$ is a subgroup of $G$, then the (left) cosets of $H$ form a set of blocks for the (left) multiplication action of $G$ on itself. So if the action is primitive, then $G$ has no nontrivial proper subgroups, so $G$ is a cyclic group of prime order. Conversely, if $G$ is cyclic of prime order, then only trivial blocks are possible since the order of a block divides the order of the group, so the action is primitive.
In general, an action of $G$ on $X$ is primitive iff the action is transitive and the stabilizer of (any or all) points is a maximal subgroup of $G$. The action of $G$ on itself has trivial stabilizer, so that action is primitive iff the trivial group is maximal. This leads to the same result.