Killing vector index manipulation

If $xi^a$ is a Killing vector then Killing's equation says that $xi_{a;b}$ is antisymmetric. It follows that, when completely contracted with a symmetric tensor, it will give zero. But $xi^a xi^b$ is a symmetric tensor. Thus I do both steps in one go. But if you want to separate the two steps, then the argument is as follows. Step 1: any second rank tensor can be written as a sum of symmetric and antisymmetric parts: $$ xi_{a;b} = xi_{(a;b)} + xi_{[a;b]} $$ so $$ xi_{mu;nu} xi^mu xi^nu = xi_{(mu;nu)}xi^mu xi^nu + xi_{[mu;nu]}xi^mu xi^nu = xi_{(mu;nu)}xi^mu xi^nu $$ where the second step is an example of the general fact that when you contract something (here $xi_{a;b}$) with a symmetric tensor (here $xi^a xi^b$), then the antisymmetric part of the first thing will not contribute (its contribution to the total will be zero). Now step 2: if $xi^a$ is a Killing vector then its covariant derivative is antisymmetric so $xi_{(a;b)} = 0$.

I think the above should settle your question about the $a_a$ equation too.