Suppose you define some equivalence relation on $X$ so that the quotient $\overline{X}$ has the property that $N\rtimes Q$ acts on $\overline{X}\times Y$ coordinatewise, with the action of $n$ on $[x]$ being $[nx]$ (here, $[x]$ is the equivalence class of $x$).
Writing ${}^qn$ for $\varphi(q)(n)$, and writing the elements of $N\rtimes Q$ as pairs $(n,q)$, we would need $[nx] = [{}^qnx]$ for all $q\in Q$ (with fixed $x$ and $n$). For $({}^qn,1) = (1,q)(n,1)(1,q^{-1})$, and looking at the actions on a pair $([x],y)$ readily gives the need for $[nx]=[{}^qnx]$.
Conversely, if $\sim$ is any equivalence relation on $x$ that satisfies this property, then the action on $\overline{X}\times Y$ given by $(n,q)([x],y) = ([nx],y)$ is readily seen to be an action:
$$\begin{align*}
(n_1,q_1)\Bigl((n_2,q_2)([x],y)\Bigr) &= (n_1,q_1)([n_2x],q_2y)\\\
&= ([n_1(n_2x)],q_1(q_2y))\\\
\Bigl((n_1,q_1)(n_2,q_2)\Bigr)([x],y) &= (n_1{}^{q_1}n_2,q_1q_2)([x],y)\\\
&= ([n_1{}^{q_1}n_2 x],(q_1q_2)y).
\end{align*}$$
The assumption on $\sim$ gives that $[{}^{q_1}n_2 x] = [n_2x]$, and hence that $[n_1{}^{q_1}n_2 x] = [n_1n_2x]$.
If you let $\sim$ be the transitive closure of the relation with $x\sim x'$ if and only if there exists $x_0\in X$, $n\in N$, and $q\in Q$ with $nx_0 = x$, ${}^qnx_0 =x'$, then any equivalence relation that is coarser than $\sim$ will do. (I may have my "coarser" and "finer" mixed up here; I mean that the equivalence classes are unions of equivalence classes under $\sim$, so that if you think of it as if it were a "topology", you would have a coarser topology, with fewer open sets).
In particular, taking $\sim$ to be the total relation ($x\sim x'$ for all $x,x'\in X$), you get the natural action on $Y$ given by mapping to the quotient (well, technically an action on $\{\bullet\}\times Y$, but that is naturally isomorphic to an action on $Y$). If the action of $Q$ on $N$ is trivial, so that ${}^qn = n$, then $\sim$ is the identity relation, so $\overline{X}=X$ and you get the natural action on $X\times Y$.
Here are a few comments and a slightly different approach, though we take advantage of some of the earlier comments. We first note the well-known (at least to people who work with factorizations) fact that if the finite group $G$ has a factorization of the form $G = AB$ with $A \cap B = 1 $ and $A,B$ subgroups, then we have $A \cap B^{g} = 1$ for all $g \in G$- for we also have $G = BA,$ and if we write $g = ba$ for some $b \in B, a \in A,$ then we have $A \cap B^{g} = A \cap B^{a} = (A \cap B)^{a} = 1.$
This means that if $G = S_{10}$ has a factorization of the form $G = AB$ with $A \cong S_{6}$ and $B \cong S_{7}$ (so that $A \cap B = 1$ as noted in the body of the question), then no non-identity element of $A$ can have the same disjoint cycle structure (in the given embedding) as any non-identity element of $B.$
Now $S_{6}$ contains commuting (and conjugate) distinct involutions which are odd permutations (in the natural representation). Hence the subgroup $A$ above contains an involution which is an even permutation in the embedding in $S_{10}.$ This is either a product of two disjoint transpositions or product of four disjoint transpositions.
It is noted in comments that $B$ must be of the form
$(S_{7} \times C_{2}) \cap A_{10},$ where the $S_{7}$ is a "natural" $S_{7}$ inside $S_{10},$ ie fixing three points. It follows that $B$ contains both involutions which are products of two disjoint transpositions, and involutions which are product of four disjoint transpositions, contrary to the remarks above.
Best Answer
Wikipedia to the rescue!
http://en.wikipedia.org/wiki/Zappa-Szep_product