Everything you wrote is correct and you have the right attitude : aggressively attacking simple examples and trying to see what is going on geometrically.
As for localization, you can use your last isomorphism if you want: localization does indeed commute with quotients.
But it is simpler to exploit your preceding isomorphism $ k[x,y]/\langle xy-1\rangle \cong k[x,x^{-1}]$ instead and to write
$$(k[x,y]/\langle xy-1\rangle)_{\langle x-1,y-1\rangle}\cong k[x,x^{-1}]_{{\langle x-1,x^{-1}-1\rangle}}=k[x,x^{-1}]_{{\langle x-1\rangle}}=k[x]_{{\langle x-1\rangle}} =k[x-1]_{{\langle x-1\rangle}}$$
Geometrical interpretation (very important!)
You are studying a hyperbola in the plane near the point $(1,1)$ by projecting it on the $x$-axis and you obtain an isomorphism with the affine line punctured at zero.
The projection sends $(1,1)$ to the point $x=1$ on the punctured line and the local ring $(k[x,y]/\langle xy-1\rangle)_{\langle x-1,y-1\rangle}$ of the hyperbola at $(1,1)$ isomorphically to the local ring of the line at $1$, namely $k[x-1]_{{\langle x-1\rangle}}$.
Note carefully that the puncture of the affine line plays no role in these local questions: the point $x=1$ only sees its immediate vicinity and doesn't care what happens at $x=0$.
The fraction ring (localization) $\rm\,S^{-1} R\,$ is, conceptually, the universal way of adjoining inverses of $\rm\,S\,$ to $\rm\,R.\,$ The simplest way to construct it is $\rm\,S^{-1} R = R[x_i]/(s_i x_i - 1).\,$ This allows one to exploit the universal properties of quotient rings and polynomial rings to quickly construct and derive the basic properties of localizations (avoiding the many tedious verifications always "left for the reader" in the more commonly presented pair approach). For details of this folklore see e.g. the exposition in section 11.1 of Rotman's Advanced Modern Algebra, or Voloch, Rings of fractions the hard way.
Likely Voloch's title is a joke - since said presentation-based method is by far the easiest approach. In fact both Rotman's and Voloch's expositions can be simplified. Namely, the only nonobvious step in this approach is computing the kernel of $\rm\, R\to S^{-1} R,\,$ for which there is a nice trick:
$\quad \begin{eqnarray}\rm n = deg\, f\quad and\quad r &=&\rm (1\!-\!sx)\,f(x) &\Rightarrow&\ \rm f(0) = r\qquad\,\ \ \ via\ \ coef\ x^0 \\
\rm\Rightarrow\ (1\!+\!sx\!+\dots+\!(sx)^n)\, r &=&\rm (1\!-\!(sx)^{n+1})\, f(x) &\Rightarrow&\ \rm f(0)\,s^{n+1}\! = 0\quad via\ \ coef\ x^{n+1} \\
& & &\Rightarrow&\ \rm\quad r\ s^{n+1} = 0
\end{eqnarray}$
Therefore, if $\rm\,s\,$ is not a zero-divisor, then $\rm\,r = 0,\,$ so $\rm\, R\to S^{-1} R\,$ is an injection.
For cultural background, for an outstanding introduction to universal ideas see George Bergman's An Invitation to General Algebra and Universal Constructions.
You might also find illuminating Paul Cohn's historical article Localization in general rings, a historical survey - as well as other papers in that volume: Ranicki, A.(ed). Noncommutative localization in algebra and topology. ICMS 2002.
Best Answer
Let us define a ring-homomorphism $u:R\to T$ by the universal property that a ring-homomorphism $f:R\to S$ factors (uniquely) through $u$ iff $f$ maps every element of $D$ to a unit and every element of $\mathfrak{a}$ to $0$. From the universal property of $D^{-1}R$, we see that any $f$ which factors through $u$ must factor through the localization map $i:R\to D^{-1}R$ via a unique map $g:D^{-1}R\to R$. Moreover, a map $f$ that factors through $i$ will factor through $u$ iff additionally $f(a)=g(i(a))=0$ for all $a\in \mathfrak{a}$. This happens iff $g$ maps all of $i(\mathfrak{a})$ to $0$. Equivalently, $g$ must factor through the quotient of $D^{-1}R$ by the ideal in $D^{-1}R$ generated by $i(\mathfrak{a})$, which is exactly $D^{-1}\mathfrak{a}$. This shows that maps factoring through $u$ are naturally in bijection with maps factoring through the composition $R\to D^{-1}R\to D^{-1}R/D^{-1}\mathfrak{a}$, so $T\cong D^{-1}R/D^{-1}\mathfrak{a}$.
On the other hand, we can do the same thing but in the reverse order to get that $T\cong \bar{D}^{-1}R/\mathfrak{a}$ as well. By the universal property of $R/\mathfrak{a}$, we see that if $f$ factors through $u$ then it factors through the quotient map $p:R\to R/\mathfrak{a}$ via a unique map $h:R/\mathfrak{a}\to S$. And a map $f$ which factors through $h$ will factor through $u$ iff additionally $f(d)=h(p(d))$ is a unit for all $d\in D$. This is equivalent to saying that $h$ factors through the localization of $R/\mathfrak{a}$ with respect to $p(D)=\bar{D}$. So $f$ factors though $u$ iff it factors through the composition $R\to R/\mathfrak{a}\to \bar{D}^{-1}R/\mathfrak{a}$.