here it says that "The complement of a pair of unlinked circles in R3 deformation retracts to S1 ∨S1 ∨S2 ∨S2 and a pair of linked circles to (S1 ×S1)∨S2. The fundamental groups are Z ∗ Z and Z × Z, respectively. "
also the answer is given here
does anyone know a place where this is actually worked out?
I am not sure I can get you exactly those deformation retractions, but I can offer some that work.
For the complement of two unlinked circles, imagine them sitting just above the $xy$ plane in $\mathbb{R}^3$. Deformation retract all of "lower $\mathbb{R}^3$" into the $xy$-plane. Cover the circles with semicylindrical sheds, as in picture $1$. Now the sheds, union the $xy$-plane take away the shadows of the sheds, is another copy of $\mathbb{R}^2$. So "upper $\mathbb{R}^3$" retracts onto this. If you view the whole situation as happening in $S^3$, then the $xy$-plane becomes a sphere and you now have a sphere with two of these shed things attached to it. This isn't quite what you wanted, but its close enough to see why $\pi_1(X)$ gets you a free group on two generators.
In my opinion the complement of two linked circles is easier. Again view the situation in $S^3$ and turn one of the circles until it goes through the point at infinity, picture 2. You can now deformation retract the complement onto a torus directly, so $\pi_1(X)$ is $\mathbb{Z}\times\mathbb{Z}$. Picture 3 shows how the retraction is happening on either side of the torus.