How to know if partial fractions have been done incorrectly?

The method you used correctly shows that if $$ \frac{2x^2-4x-4}{(x-1)(x-3)} = \frac{A}{x-1} + \frac{B}{x-3} $$ then it must be the case that $A = 3$ and $B = 1$. So this shows that those values of $A$ and $B$ are the only ones that have any chance of working; it doesn't show that those values actually work.

It is clear that the equality $2x^2-4x-4=A(x-3)+B(x-1)$ isn't an identity as the coefficients corresponding to the term of the degree $2$ aren't the same (you have 2 on the left-hand side and $0$ on the right hand side).

When you set $x=1$ and $x=3$ and you solve in $A$ and $B$, you find the values of $A$ and $B$ for which the equation $2x^2-4x-4=A(x-3)+B(x-1)$ has $1$ and $3$ as solutions. Geometrically, you find the only line of the family $y=A(x-3)+B(x-1)$ with intersects the parable $y=2x^2-4x-4$ at the points $(1,-6)$ and $(3,2)$. But obviously, the fact that the line and the parable have two points in common is not enough to say that they $\textit{coincide}$.

When you introduce the third constant $C$, you get an equation of the form $2x^2-4x-4=A(x-3)+B(x-1)+C(x-3)(x-1)$. Then you take 3 values of $x$ and you find $A,B,C$ such that the relations hold. In that case, you are forcing two parables to have 3 common points and it is a general fact that two $n$-th degree polynomials with $n+1$ points in common are the same, so your initial expression becomes an identity for the values of $A,B,C$ that you found.