The correct way to think about adjoining multiplicative inverses to elements already in the polynomial is localization.
What you are basically doing when adjoining $1/2$ is that you are declaring that $2$ ought to be a unit in $R[1/2]$ and that the rest of the ring ought to be have as close as possible to $R$ (formally, it satisfies a certain universal property; see the wiki for more detail).
In any case, let $S$ be the multiplicative set (monoid really) generated by $2$ in $R$ (in this particular case $S=\{1,2,4,8\}$; we require $1\in S$ for the following to work). Then if $2$ is a unit, units should also be $4$ and $8$. The key property about units is that they are invertible, so all the elements of $R[1/2]$ will be of the form $r/s$ where $r\in R$ and $s\in S$. Just like regular fractions, however, sometimes two elements $r/s$ and $r'/s'$ are actually the same element. Having $r/s=r'/s'$ would imply that $s'r-rs'=0$ in $R[1/2]$.
Multiplication in $R[1/2]$ is not the same as multiplication in $R$ for the following reason. Suppose that in $R$ we have $ts=0$ for some $s\in S$. Then in $R[1/2]$ we should still have $ts=0$, but now, because in $R[1/2]$ we have that $s$ is unit, this implies that $t=0$! In other words, the map $R\to R[1/2]$ given by $r\to r/1$ is not injective since anything that annihilates something in $S$ must now be sent to $0$.
This allows us to write down explicitly the condition of $r/s=r'/s'$ as an equation in $R$: there exists an element $s_0\in S$ annihilated by $(s'r-rs')$, i.e. *$(s'r-rs')s_0=0$ in $R$*. It turns out that this condition is sufficient to describe all of $R[1/2]$.
Now, in the case where $R=\mathbb Z$, $S=\{2^k\colon k\geq 0\}$, $(s'r-rs')s_0=0$ in $R$ implies $s'r-rs'=0$ since $S$ consists only of non-zero divisors (hence we can cancel by them) in $\mathbb Z$, and so $R[1/2]$ is just the ring of the dyadic fractions, that is, fractions with denominator that is a power of $2$.
How does the above answer your question? Well, looking at $R[x]/\left<2x-1\right>$, you must consider the kernel of the map $R\to R[x]/\left<2x-1\right>$. The above gives you intuition that these should be things killed by $2,4$, or $8$ which are $3$, $6$, and $9$, so you expect the kernel to be $\{0,3,6,9\}$.
Lo and behold: it is, since really, that ideal is $R\cap \left<2x-1\right>$ and the only way that a multiple of $2x-1$ is equal to a constant is if all of the coefficients on indeterminates $x$ are $0$. Hence, if $\sum a_ix^i(2x-1)=-a_0$, then we need $2a_i-a_{i+1}=0$, or equivalently $2a_i=a_{i+1}$, which implies that $2^na_0=0$, establishing that $a_0$ is in fact inside the ideal $\{0,3,6,9\}$. That each of $0,3,6,9$ can be achieved is evident from multiplying $2x-1$ by $-\sum 2^ix^ia_0$.
This generalizes to show that in general the kernel of $R\to R[x]/\left<sx-1\right>_{s\in S}$ for a multiplicative set $S$ is the ideal generated by elements that kill at least one element of $S$.
It follows that $R[x]/\left<sx-1\right>_{s\in S}\cong R/I\left<sx-1\right>_{s\in S\setminus I}$, that is, inverting the multiplicative set $S$ in $R$ is the same as inverting the image of the multiplicative set $S$ in $R/I$, which image will have no such ideal. So we end up with showing that if $R\to R[x]/\left<sx-1\right>$ is injective, then the latter is isomorphic to fractions $r/s$ with $r/s=r'/s'$ if $rs'-r's=0$ since the image of $S$ has no zero-divisors, which you can do on your own.
Finally, applying the above to your question, we see that $\mathbb Z/12/\left<2x-1\right>\cong\mathbb Z/3/\left<2x-1\right>\cong\mathbb Z/3$, the latter holding because $2$ is already a unit in $\mathbb Z/3$.
