What does it mean for localization to be same if multiplicative set is different? $T^{-1}R = S^{-1}R$? If R is a ring and p is a prime ideal, I was told that somehow $Frac(R/p)$ was the localisation $(R \setminus p)^{-1}R/p$. I thought it was more like the localisation $(R/p \setminus\{p\})^{-1}R/p$ but apparently these are the same. But why? What exactly should I verify? (I am new to all this localization business).
Also, in general what does it mean for $T^{-1}R$ and $S^{-1}R$ to be the same if $T$ and $S$ are different?

Maybe $(R\setminus p)^{-1}(R/p)$ is confusing since $R\setminus p$ is not a subset of $R/p$? It would indeed be better to write $(\overline{R\setminus p})^{-1}(R/p)$, where the overline means taking classes modulo $p$. Alternatively, the latter can be written as $((R/p)\setminus\{\bar 0\})^{-1}(R/p)$ which coïncides with your suggestion.
Your question as to when two localizations $T^{-1}R$ and $S^{-1}R$ are isomorphic is a nice exercise: let $\tilde S$ be the saturation of a multiplicative subset $S$ of $R$ defined by
$$
\tilde S=\{r\in R\mid \exists t\in R\colon rt\in S\}.
$$
In words, $\tilde S$ is the subset of all divisors of the elements of $S$.
One can prove that $\tilde S$ is again multiplicative and that the natural morphism from $S^{-1}R$ to $\tilde S^{-1} R$ is an isomorphism. Also, $\tilde S$ is exactly the subset of elements of $R$ whose image in $S^{-1}R$ is invertible. With that in hand, it is easy to prove that the localizations $S^{-1}R$ and $T^{-1}R$ are isomorphic if and only if $\tilde S=\tilde T$. One has to note that by isomorphism of localizations I mean here that there is an isomorphism between $S^{-1}R$ and $T^{-1}R$ that is the identity on the image of $R$. Without that precision, it may well be that the rings $S^{-1}R$ and $T^{-1}R$ are abstractly isomorphic without $\tilde S=\tilde T$. An example of the latter are the rings $\mathbf Z[X,Y]_X$ and $\mathbf Z[X,Y]_Y$, where $R_r$ denotes the localization of the ring $R$ with respect to the multiplicative subset $\{1,r,r^2,\ldots\}$.
