Universal Coefficient Theorem 1.17. Let $A\in\mathcal N$. Then there is a short exact sequence
$$ 0\to \text{Ext}(K_*(A),K_*(B))\stackrel{\delta}\to KK_*(A,B)\stackrel{\gamma}\to \text{Hom}(K_*(A),K_*(B))\to 0$$
The map $\gamma$ has degree $0$ and the map $\delta$ has degree $1$.
This is a theorem from THE KUNNETH THEOREM AND THE UNIVERSAL COEFFICIENT THEOREM FOR KASPAROV’S GENERALIZED K-FUNCTOR, JONATHAN ROSENBERG AND CLAUDE SCHOCHET
I can't figure out whether $\delta$ is a $\text{Ext}(K_{i+1}(A),K_{i+1}(B))\to KK_i(A,B)$ map or a $\text{Ext}(K_{i}(A),K_{i}(B))\oplus \text{Ext}(K_{i+1}(A),K_{i+1}(B))\to KK_i(A,B)$ map or a $\text{Ext}(K_{i}(A),K_{i+1}(B))\oplus \text{Ext}(K_{i+1}(A),K_{i}(B))\to KK_i(A,B)$ map.
It seems the paper is proving the first one but I am not sure.
I still don't know what degree means exactly, but I think I have figured out what map it is here.
It is $$ 0\to \oplus_j \text{Ext}(K_j(A),K_{i+j-1}(B))\stackrel{\delta}\to KK_i(A,B)\stackrel{\gamma}\to \oplus_j \text{Hom}(K_j(A),K_{j+i}(B))\to 0 $$