I am too lazy to calculate the numbers of elements with $k$ cycles in $S_8$ but if you do that yourself a solution could work as follows. (I will use this version of Redfield-Polya and $[n]$ shall denote $\{1,\dots,n\}$.)
Let us take $X = [13]$ the set of fruits, where $G= S_5 \times S_8$ acts on $X$ such that the first five apples and the later eight oranges are indistinguishable. Then
$$K_n = |[n]^X/G|$$
is the number of ways to distribute these apples and oranges among six distinguishable children. And
$$ N_n = K_n -n\cdot K_{n-1}$$
the of ways to distribute these apples and oranges among six distinguishable children such that every child must recieve at least one piece of fruit.
Now by the Theorem
$$K_n = \frac{1}{|G|} \sum_{g\in G} n^{c(g)} = \frac{1}{5!\cdot 8!} \left(\sum_{g\in S_5} n^{c(g)}\right)\left(\sum_{g\in S_8} n^{c(g)}\right) = \frac{1}{5!\cdot 8!} \left(\sum_{i\in [2]} d_i n^{i}\right) \left(\sum_{i\in [4]} e_i n^{i}\right),$$
where $c(g)$ is the number of cycles of $g$, $d_i$ the number of permutations of $S_5$ with exactly $i$ cycles and $e_i$ the number of permutations of $S_8$ with exactly $i$ cycles.
The number that we are looking for in the end is $N_6$.
If you have $9$ apples, you can choose $0, 1, 2, 3, ... 9$ of them. So there are $10$ ways to choose. Similarly for oranges, you have up to $7$ ways to choose. By the fundamental counting principle, you have $10\times7 = 70$ possible ways of choosing fruits (or no fruit). There is only one way to choose no fruit so we subtract $1$.
If you've used generating functions, you could also see it but it would also be a hassle to expand.
Also, have you ever encountered a problem such as "How many divisors does x have? (x non-prime)" It is similar to that where you add $+1$ to each power of the prime factorization when multiplying them.
If each apple and each orange were distinguishable, there would again be only one way to choosing nothing.
If the $9$ apples were distinguishable and the $6$ oranges were distinguishable, I believe this is the same as considering $9 + 6 = 15$ different objects.
How many ways are there to choose from $15$ different objects? There are $2^{15}$ such ways (including not choosing anything). Then $2^{15}-1$ is gives you the number of ways in which you have to choose at least one fruit. This is analogous to a power set.
Best Answer
Right, if there were $13$ oranges, there are $66$ possibilities.
Let's make use of this: We first distribute $13$ oranges, and then we exchange one orange for an apple. There are $66$ ways to distribute the oranges, and $3$ ways to select a child which will exchange an orange for an apple. So in total, there are $$66\cdot 3 = 198$$ possibilities.