How to compute localizations of quotients of polynomial rings

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 $x=0$.