Hi there! The point is that the scalar quantity you formed is constant along the geodesic! Using your notation, [itex]\triangledown _{U}(X_{i}U^{i}) = U^{j}\triangledown _{j}(X_{i}U^{i}) = U^{j}U^{i}\triangledown _{j}X_{i} + X_{i}U^{j}\triangledown _{j}U^{i}[/itex]. Note that [itex]U^{j}U^{i}\triangledown _{j}X_{i}[/itex] vanishes because [itex]U^{j}U^{i}[/itex] is symmetric in the two indices whereas, by definition of a killing vector, [itex]\triangledown _{j}X_{i}[/itex] is anti  symmetric in the two indices and it is very easy to show that the contraction of a symmetric tensor with an anti  symmetric one will vanish. The second term [itex]X_{i}U^{j}\triangledown _{j}U^{i}[/itex] vanishes simply because U is the tangent vector to a geodesic thus we have that [itex]\triangledown _{U}(X_{i}U^{i}) = 0[/itex]. In particular note that if this geodesic is the worldline of some freely falling massive particle then its 4  velocity is the tangent vector to the worldline and we can re  express the condition for the worldline being a geodesic in terms of the 4  momentum of the particle (and for photons just define the geodesic condition like this) and we can have that if [itex]X^{i}[/itex] is a killing field on the space  time then [itex]X_{i}P^{i}[/itex] will be constant along this geodesic. It is a geometric way of expressing local conservation of components of the 4  momentum; these killing fields are differentiable symmetries of the space  time and you might be able to see that more clearly by the fact that the lie derivative of the metric tensor along the killing field will vanish.
