Let’s try to work in $\Bbb Z$, at least at first: it’s the most familiar additive group.
Clearly $W(A,B)\supseteq\{\langle a,a,b,b\rangle:\langle a,b\rangle\in A\times B\}$, so $|W(A,B)|\ge N^2$. Thus, to get $|W(A,B)|$ comparable to $N^2$, we need to try to arrange matters so that $a+b=a'+b'$ iff $\langle a,b\rangle=\langle a',b'\rangle$. Clearly that’s as small as we can get.
Suppose, on the other hand, that for each $\langle a,b\rangle\in A\times B$ and each $a'\in A$, $a+b-a'\in B$; then each $\langle a,b\rangle\in A\times B$ would generate $N$ members of $W(A,B)$, and $|W(A,B)|$ would be comparable to $N^3$. It’s not too hard to see that this is as large as we can get.
For the first problem, then, we’d like to choose $A,B$, and $C$ so that
for each $\langle a,b\rangle\in A\times B$ and $a'\in A$, $a+b-a'\in B$ iff $a'=a$;
for each $\langle a,c\rangle\in A\times C$ and $a'\in A$, $a+c-a'\in C$ iff $a'=a$; and
for each $\langle b,c\rangle\in B\times C$ and $b'\in B$, $b+c-b'\in C$.
Suppose that $A=\{1,\dots,N\}$. Then for any $a,a'\in A$ we have $-N. Suppose now that we can choose $B$ so that if $b$ and $b'$ are distinct members of $B$, then $|b-b'|\ge N$. Now suppose that $\langle a,b\rangle\in A\times B$ and $a'\in A$. If $a+b-a'=b'\in B$, then $a-a'=b'-b$; but $|a-a'|, and $|b'-b|\ge N$ unless $b=b'$, in which case $a=a'$. Thus, such a $B$ would satisfy (1). Is there such a $B$? Sure: just let $B=\{kN:k=1,\dots,N\}$.
If we let $C=B$, (2) is automatically satisfied. (3) isn’t satisfied, but it is true that if $\langle b,c\rangle\in B\times C$, $b'\in B$, and $b+c>b'$, then $b+c-b'\in C$. How many triples $\langle b,c,b'\rangle\in B\times C\times B$ are there such that $b+c>b'$?
It’s actually easier to count the triples with $b+c\le b'$. I leave it to you to verify that there are
$$\begin{align*} \sum_{i=1}^{N-1}i(N-i)&=N\sum_{i=1}^{N-1}i-\sum_{i=1}^{N-1}i^2\\ &=\frac12N^2(N-1)-\frac16N(N-1)(2N-1)\\ &=\frac{N(N-1)}6\big(3N-(2N-1)\big)\\ &=\frac16(N^3-N) \end{align*}$$
of these, and therefore $\frac56N^3+\frac16N$ triples with $b+c>b'$. This should qualify as comparable to $N^3$.
Edit: For the second problem, suppose that $N=2n$ and try
$$\begin{align*} A&=\{1,\dots,n\}\cup\{kn^2:k=1,\dots,n\},\\ B&=\{1,\dots,n\}\cup\{kn^4:k=1,\dots,n\},\text{ and}\\ C&=\{kn^2:k=1,\dots,n\}\cup\{kn^8:k=1,\dots,n\}\;. \end{align*}$$
Try to show that the overlapping parts of $A$ and $B$ and of $A$ and $C$ ensure that $|W(A,B)|$ and $|W(A,C)|$ are at least some fixed multiple of $N^3$, while $B$ and $C$ are related more like my $A$ and $B$ in the first problem.