$H^0$ is a cohomological functor

Question

I was reading Section 1.5 of Sheaves on Manifolds by Kashiwara and Shapira. In the proof of Propositon 1.5.6, they states that in order to prove the functor $H^0:K(\mathcal{C})\rightarrow \mathcal{C}$ is a cohomological functor, where $\mathcal{C}$ is an abelian category, it suffices to prove that $H^0(Y)\rightarrow H^0(M(f))\rightarrow H^0(X[1])$ exact. I failed to see why. It is clear that the condition is necessary since $Y\rightarrow M(f)\rightarrow X[1]\rightarrow Y[1]$ is a distinguished triangle. But why is it sufficient?