You can never get answers with $\frac{\frac{a}{b}}{\frac{c}{0}}$ where $a, b, c$ $\ne$ $0$.
$\frac{a}{c}$$\ne$ $0$ and $\frac{b}{0}$ $=$ $undefined$.
But why is there a specific answer ($0$ definitely) if $\frac{a}{b}\cdot\frac{0}{c}$?
You can never get answers with $\frac{\frac{a}{b}}{\frac{c}{0}}$ where $a, b, c$ $\ne$ $0$.
$\frac{a}{c}$$\ne$ $0$ and $\frac{b}{0}$ $=$ $undefined$.
But why is there a specific answer ($0$ definitely) if $\frac{a}{b}\cdot\frac{0}{c}$?
The expression $\frac{a/b}{c/0}$ means:
Of course you can immediately see the problem with that: the second step is not possible to do. So this expression doesn't give a well-defined number.
However, it is true that if $b,c,d$ are all nonzero numbers, then $\frac{a/b}{c/d}$ is a well-defined number, and this number is equal to $\frac{a}{b} \cdot \frac{d}{c}$. And it is true that this last expression is well-defined when $d = 0$. However this rule can only apply when $d \neq 0$; it doesn't mean that $\frac{a/b}{c/0}$ becomes well-defined.
Here is an example of why you need to be careful. You certainly agree that $\frac{xy}{y} = x$, right? Now take $x = 2$ and $y = 0$, and you get $2 = \frac{2 \times 0}{0}$. But of course $2 \times 0 = 0$, so $2 = \frac{0}{0}$. Now I can apply the same "rule" to $3 = \frac{3 \times 0}{0} = \frac{0}{0}$, and I finally get $2 = 3$. Which is obviously nonsense.
What went wrong? Well, the "rule" $\frac{xy}{y} = y$ is only valid when $y \neq 0$, because otherwise you're dividing by zero! You can't just blindly apply "rules" without checking that the conditions for these rules are satisfied, and you can't act like undefined expressions (like $\frac{c}{0}$) mean something, otherwise you will run into trouble. The tag fake-proofs is full of such examples.