Does this imply that when an observer which is at rest at some point of the spacetime measures a quantity I should use a locally Minkowskian coordinate system (tangent space of the point P of the observer on the Manifold) and in that coordinate system the metric is not independent of time, since he sees that this quantity changes according to the point of spacetime he measures the quantity from?
I haven't read Schutz, but from reading your question it sounds like his presentation of this topic has some deficiencies, and your confusion may be natural given those deficiencies. He's discussing this in terms of the (non-covariant) derivative of the metric with respect to a coordinate, which immediately creates some serious problems. That quantity simply isn't measurable. If you want to measure the metric or its derivatives, you end up with the following restrictions:
A local observer can't measure the metric. (This is for the same reason that you can't measure an absolute potential energy. The metric plays a role in GR analogous to that of the potential in Newtonian gravity.)
A local observer can't measure the derivative of the metric. (That derivative would basically be the gravitational field, which is not measurable because of the equivalence principle.)
A local observer can measure the second derivative of the metric, which is essentially a measure of tidal stresses.
So Schutz's presentation makes use of a derivative that has no physical interpretation.
Yes, any time any observer measures a vector quantity (such as the energy-momentum vector), they are implicitly doing so in some local Minkowski frame. (They could, for example, measure the vector's inner product with some other vector, but then they are effectively using this other vector as a coordinate axis of some Minkowski frame.)
Will any observer ever see this quantity conserved when he measures it or is it just a mathematical construct?
To verify this conservation law, the observer needs to have global information, not just local information. Basically they need to measure the quantity $E^*$ in a local static frame, then use their global knowledge (of the metric and of their position in the spacetime) to determine $E$. They can then verify that $E$ is conserved.
The inability to determine such a conservation law based on purely local information is baked in to the structure of GR. Energy-momentum is a vector, and you can't compare vectors at different points in spacetime except by parallel transport. You can certainly verify that a test particle's energy-momentum vector is preserved under parallel transport along its own geodesic of motion, but you end up with a triviality, which is essentially that the test particle had the same free-fall motion that you did. This is just a test of the equivalence principle, and it holds even in a spacetime that does not have any symmetry.
To resolve the issues you're talking about in a more satisfactory way than in Schutz's presentation, you really need to use the notion of a Killing vector.
Best Answer
Let $V^{\mu}=\frac{dx^\mu}{d\lambda}$. Then, $\epsilon=-V_\mu V^\mu$. To see the conservation of this quantity along the geodesic we have to look at the covariant derivative along the curve, that is $$V^\nu\nabla_\nu\epsilon=-V^\nu\nabla_\nu(V_\mu V^\mu)=-V^\nu V^\mu\nabla_\nu V_\mu -V^\nu V_\mu\nabla_\nu V^\mu.$$ But metric compatibility implies that $\nabla_\nu V_\mu=g_{\mu \sigma}\nabla_\nu V^\sigma$. Also using the geodesic equation $V^\nu \nabla_\nu V^\mu=0$, we find $$V^\nu\nabla_\nu\epsilon=0,$$ which shows that $\epsilon$ is a conserved quantity.