There are two linear maps $m:V \otimes V \rightarrow V$ and $\Delta:V \rightarrow V\otimes V$ in the definition of the differential of Khovanov homology. So my question is why do they map elements as below?
$$m:v_+ \otimes v_- \rightarrow v_-$$ $$m:v_+ \otimes v_+ \rightarrow v_+$$ $$m:v_- \otimes v_+ \rightarrow v_-$$ $$m:v_- \otimes v_- \rightarrow 0$$
and
$$\Delta:v_+ \rightarrow v_+ \otimes v_-+v_-\otimes v_+$$ $$\Delta:v_- \rightarrow v_- \otimes v_-$$
Thank you!
First note that there are other definitions you could choose. Eun Soo Lee uses an altered differential (introduced in section 4 of this paper), and the resulting theory was used to great effect by Jake Rasmussen in his paper on slice genus.
For discussion of Khovanov's differential, I recommend Bar-Natan's explanation in his colorful introduction to Khovanov homology. Paraphrasing section 3.2, the differential ought to be of degree 0 and be invariant under any reordering of the cycles. This requires that the multiplication and comultiplication maps be of degree $-1$ and be commutative and co-commutative, respectively. Working in $\mathbb{Z}/2\mathbb{Z}$, this determines $m$ and $\Delta$.