Edited: Suppose we are given two disjoint permutations $(a_1,\dots,a_m)$ and $(b_1,\dots,b_n)$. I want to prove that (if it is correct)
$$\text{sgn}(a_1,\dots,a_m, b_1,\dots,b_n)\times \text{sgn}(b_1,\dots,b_n, a_1,\dots,a_m)=(-1)^{\sum_{i=1}^m (a_i-1)+\sum_{i=1}^n (b_i-1)}.$$
Where $\text{sgn}$ denotes the sign of permutation which is $+1$ or $-1$ depending on whether the permutation is even or
odd.
I know that parity of a $k$-cycle is $k-1$. I think $(a_1,\dots,a_m, b_1,\dots,b_n)$ is same as $(a_1,\dots,a_m)(b_1,\dots,b_n)$ because they are disjoint cycles, so $\text{sgn}(a_1,\dots,a_m, b_1,\dots,b_n)=\text{sgn}(a_1,\dots,a_m)\cdot\text{sgn}(b_1,\dots,b_n)$. Right? And $\text{sgn}(a_1,\dots,a_m)=(-1)^{m-1}$ and $\text{sgn}(b_1,\dots,b_n)=(-1)^{n-1}$ But these seems do not imply the above claim.
Any suggestion?
Firstly I think from looking at the page of the book you reference in a comment to the answer by Henno Brandsma (which answers the question you originally posed) that what is meant here by $\text{sgn}(c_1,c_2,\dots, c_N)$ is the sign of the permutation $i\mapsto c_i$.
What you are looking at is the case where $N=m+n$, and the first $m$ $c_i$ are $c_i=a_i$ and the last $n$ $c_i$ are $c_{i-m}=b_i$.
When you look at $$ \text{sgn}(a_1,\dots,a_m, b_1,\dots,b_n)\times \text{sgn}(b_1,\dots,b_n, a_1,\dots,a_m) $$ you might as well ask for $$ \text{sgn}(a_1,\dots,a_m, b_1,\dots,b_n)^{-1}\times \text{sgn}(b_1,\dots,b_n, a_1,\dots,a_m) $$ since for every permutation $\pi$ we know that $\pi^1$ and $\pi$ have the same sign.
We also know that for any pair of permutations $\pi_1,\pi_2$ the sign of $\pi_1\pi_2$ is just the product of the signs of $\pi_1$ and $\pi_2$. So what you really want to calculate is $$ \text{sgn}(a_1,\dots,a_m, b_1,\dots,b_n)^{-1} (b_1,\dots,b_n, a_1,\dots,a_m). $$
But what is this permutation? It is the permutation taking $a_1$ to $1$ to $b_1$, $a_2$ to $2$ to $b_2$, ..., $b_n$ to $n$ to $a_m$. [To avoid confusion I read my permutations left to right.]
So how many swaps does it take to move the $a_i$ behind the $b_i$? Well each of the $m$ $a_i$ must slip past $n$ of the $b_j$, so that there are $mn$ swaps.
Hence the sign of the permutation is $(-1)^{mn}$ or if you prefer $(-1)^{m(N-m)}$; the latter is the form given in "Solution (2)" of your book.
Edit: Further question in the comments below.
You are asking (I think) where one of the intermediate formulas in Solution(2) of the book comes from. At this stage we are trying to compute the sign of the permutation $(a_1,a_2,\dots, a_m,b_1,\dots, b_n)$ where this notation still means the permutation taking $1\dots (m+n)$ to these values.
I believe that in the book we have this added information: $a_1<a_2<\dots<a_m$ and $b_1<b_2<\dots<b_n$. To calculate the sign of this permutation note that we have to first swap the first $(a_1-1)$ elements $b_j$ past the $m$ elements $a_i$. Next we have to swap the next $(a_2-a_1-1)$ elements $b_j$ past $(m-1)$ elements $a_i$. Next we have to swap the next $(a_3-a_2-1)$ elements $b_j$ past $(m-2)$ elements $a_i$. And so on. When I calculate this sum and simplify it I get -f I have made no mistake- $\sum_{i=1}^{m} a_i - {m\choose 2}$ which is what is required in the book.