If we localize $A = \mathbb{Z} / 6\mathbb{Z}$ at the ideal $(2)$, we get the ring $A_{(2)}$ containing fractions $n / s$ for $n \in A$ and $s \in S = \{1, 3, 5\}$. Because $2$ and $4$ are killed by $3$, we have $n / s = 0$ if $n \in \{2, 4\}$. In addition, we have $3 / 1 = 1 / 1 + 2 / 1 = 1$ and $5 / 1 = 3 / 1 + 2 / 1 = 1$, so the only elements of $A_{(2)}$ are $0$ and $1$. That is, $A_{(2)} \cong \mathbb{Z}/2\mathbb{Z}$.
The computation of $A_{(2)}$ is easy enough, but I don't have any intuition why this is true. It doesn't seem obvious to me that "formally adding odd denominators to elements of $\mathbb{Z}/6\mathbb{Z}$" should reduce it down to a ring with two elements. Is there a nice way of thinking about this example, or is it somewhat pathological?
The general idea is that if $ab=0$ and $a$ is invertible, then $b=0$.
In $\mathbb{Z}/6\mathbb{Z}$, we have $3 \cdot 2 = 0$, so if your localization inverts 3, then this forces 2=0 and you get $\mathbb{Z}/2\mathbb{Z}$.
Why don't we get anything extra beyond $\mathbb{Z}/2\mathbb{Z}$? Because in $\mathbb{Z}/2\mathbb{Z}$, the (images of the) odd numbers from the original $\mathbb{Z}/6\mathbb{Z}$ are already invertible, so you don't need anything else. This follows from the universal property of the localization. Intuitively, it is the "simplest" ring $R$ with a map $\mathbb{Z}/6\mathbb{Z} \to R$ such that images of all the odd numbers are invertible.
Similar types of intuitive arguments are available in other contexts. For example, if $R$ is a ring then the quotient $R[x]/(x)$ is isomorphic to $R$. The quotient by the ideal $(x)$ forces $x=0$, leaving you with $R$. Why don't you get anything beyond $R$? Because the quotient is the simplest ring with a map from $R[x]$ sending $x$ to 0.
The precise meaning of "simplest" is given by the universal property.