Definition Given disjoint manifolds $M$, $N\subset \Bbb{R}^{k+1}$, the linking map $\lambda: M\times N\to S^k$ is defined by $\lambda(p, q) = (p - q)/||p - q||$. If $M$ and $N$ are compact, oriented, and boundaryless, with total dimension $m + n = k$, then the degree of $\lambda$ is called the linking number $l(M, N)$.
Now let $K$ be the unit circle in the $xy$-plane with center (-1,0,0), oriented clockwise and let $M$ be the unit circle in the $xy$-plane with center $(2,0,0)$ oriented clockwise. How do I explicitly compute $l(K,M)$?
Since these circles are not "linked" the result must be $0$. First, for a regular value $z$ of $\lambda$ I need to determine the number of elements of $\lambda^{-1}(z)$. Then I should compute the Jacobian matrix of $\lambda$, BUT it is not a square matrix? So how do I obtain the sign of its determinant? I don't even know how many points are there in $\lambda^{-1}(z)$. Please help.
In your example, where $K,M$ are disjoint circles in the $x,y$ plane of $\mathbb{R}^3$, the vector $\lambda(p,q)$ is parallel to the $x,y$ plane, and therefore $\lambda(p,q)$ is a point on the equator of $S^2$. So, $\lambda :K \times M \to S^2$ is a map from the 2-torus $T^2$ to $S^2$ whose image is contained in the equator. This map is not surjective. You may compute the degree using any point on the range $S^2$, so feel free to choose a point $z$ which is not in the image of the map. Since $\lambda^{-1}(z)$ is empty, $degree(\lambda)=l(K,M)$ is therefore equal to zero.