Therefore, could we consider N*G as the zero ?
It is not zero, it is called infinity point. Essentially to compute
k*G if
k%N = 0 then we get point at infinity.
(-k*G+k*G)= N*G only for (k=1).
That is true for all k values and the result is point at infinity.
-k*G + k*G = (N-k)*G + k*G = (N-k+k)*G = N*G = 0*G = infinity
I'm not sure about that, we can also write:
-1*G + 1*G = N*G
-2*G + 2*G = (-1*G - 1*G) + (1*G +1*G) = 2N*G
...
-k*G + k*G = N*K*G
Therefore, there is probably a different "infinity" point for each pair (-k*G, k*G), which is in accordance with the fact that addition of x-axis symetrical points forms parallel lines.
However, in reality with k=2, we get a point not equal to 2N*G and with no apparent link to N*G :
-2*G + 2*G=(49667982834466148699028630885550314015746939561788444090107179747772288677390, 37758011528734597361759657276645959964664126776856991748971785837209648797521).
So for me the question still needs clarifications.
Before generalizing the formula to any K<N, let's just understand what happens with k=2 that is different from k=1.