On page 72 of Eisenbud's text on Commutative Algebra with a view $\rightarrow$ A.G. he writes that $(\mathbb Z/(12))_{(2)}=(\mathbb Z/(4))_{(2)}=\mathbb Z/(4)$. I'm having trouble understanding the meaning of this. Are the equalities supposed to be isomorphisms?
What is the image of say, the element $ 6/7\in (\mathbb Z/(12))_{(2)}$ in $(\mathbb Z/(4))_{(2)}$ under such an isomorphism?
These should be isomorphisms. Already in $\Bbb Z/(12)$, $6/7=6$ so in its localisation at $(2)$, $6/7=2$ as $6\equiv 2\pmod4$.
How does this localization work? To get $(\Bbb Z/(12))_{(2)}$ we take $\Bbb Z/(12)$ and force all elements outside $(2)$ to have an inverse. In particular $3$ must gain an inverse. In $\Bbb Z/(12)$, $3\times 4=0$ so in $\Bbb Z/(12)_{(2)}$ $4$ becomes zero and $3$ becomes the unit $-1$. This means that $\Bbb Z/(12)_{(2)}$ is a quotient of $\Bbb Z/(4)$.
Consider the natural map $\phi :\Bbb Z/(12)\to \Bbb Z/(4)$. All odd numbers in $\Bbb Z/(12)$ are mapped by $\phi$ to $\pm 1$ in $\Bbb Z/(4)$ which are units in $\Bbb Z/(4)$. By the universal property of localisation, this shows that $\Bbb Z/(4)\cong \Bbb Z/(12)_{(2)}$.
