Let consider the ring $A:=k[x_1,\dotsc,x_n]$ and the ideal $I:= (f_1,\dotsc,f_r)$. My point of interest is following conclusion about Kähler differentials:
I know that $\Omega^1_{k[x_1,\dotsc,x_n]}$ is a free $k[x_1,\dotsc,x_n]$-module with basis $d(x_1),\dotsc,d(x_n)$
Futhermore for any ideal $I \subseteq A$ there was proved that the sequence $I/I^2 \xrightarrow{d} \Omega^1_{A} \otimes_A A/I \to \Omega^1_{A/I} \to 0$ is exact.
Now the question: Why the both facts above already imply that $\Omega^1_{k[x_1,\dotsc,x_n]/(f_1,\dotsc,f_r)}$ is free as module over $k[x_1,\dotsc,x_n]/(f_1,\dotsc,f_r)$?
What has it to do with abstract definition $\Omega^1=I/I^2$ for differential forms?
If you have a free module $F$ over a ring $R$ and $I$ an ideal, then it is immediate to see that $F\otimes_R R/I=F/IF$ is a free module over $R/I$ (with the basis chosen as images of the basis of $F$), which was your first question. For, the latter, the ideal $I$ has a different meaning here (you may remember, this comes from looking at the diagonal, and so check carefully what it means).