One has ζd∘id=0 by Lemma 1. For every p∈Spec(A) such that dim(A/p)=d, one has ζd([A/p]≤d)=zd(A/p)=[V(p)], hence the homomorphism ζd is surjective. By IV, § 1, No. 4, Theorem 1, K(C≤d) is generated by the [A/p]≤d, where p∈Spec(A) and dim(A/p)≤d; consequently, every element ξ of K(C≤d) can be written ξ=id(η)+∑i=1kni[A/pi]≤d, with η∈K(C≤d−1), ni∈Z and dim(A/pi)=d for 1≤i≤k; one has ζd(ξ)=∑i=1kni[V(pi)] and consequently ζd(ξ)=0 implies ξ=id(η)∈Im(id), whence Ker(ζd)=Im(id).
One reasons analogously for the second sequence.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.