Why does instrument exogeneity imply conditional mean zero?
No, you're right, $Cov[Z_i,u_i]=0$ does not imply $E[u_i|Z_i]=0$ in general. If the author defined "instrument exogeneity" as $Cov(Z_i,u_i)=0$ previously, he/she is being careless here.

"Instrument exogeneity" is often defined as $E[u_i|Z_i]=0$ too, in which case all you need to do would be to understand "implies" as "means". But since exogeneity is defined as uncorrelatedness (zero covariance), this issue has to be addressed. the author'd better fix it in a later version. How about "Instrument exogeneity usually means ..." instead of "Instrument exogeneity implies ..."?