How do I calculate the (apparent) gravitational pull with General Relativity? Assume a static metric with (known) components $g_{\mu\nu}$. I'd like to know what is the gravitational pull $g$ of a test particle placed on an arbitrary point $X$.
The gravitational pull being defined as the acceleration the particle suffers as measured by an observer sitting in a reference frame fixed at the origin. What are (theoretically speaking) the steps one needs to take to find this acceleration? (no need to actually calculate for the general case, just list the general steps)
Furthermore, what is the result for the simple case of the Schwarzschild metric?
In other words, what is the gravitational pull of a star or black hole? Will it matter if the test particle is moving or is static?
 A: This problem is somewhat ill-formed.  It's difficult to even define what 'static at the origin' means--static relative to what?  What I will do is show the steps one needs to take to derive the motion of a particle, however.  The following will be more constructive than explanatory, as explaining all of this thoroughly would be close to a whole chapter in a relativity/differential geometry textbook.  
Given the metric tensor $g_{ab}$, one can define the Christoffel symbols $\Gamma_{ab}{}^{c}$ by the equation$^{1}$ $\Gamma_{ab}{}^{c} = \frac{1}{2}g^{cd}\left(\partial_{a}g_{bd}+\partial_{b}g_{ad}-\partial_{d}g_{ab}\right)$.  While the Christoffel symbols serve an array of purposes in relativity, the simplest notion that one can tie to them is that they define a notion of parallel transport in the spacetime--a vector $v^{a}$ is parallel transported along a curve $x^{a}$ if $x^{a} \nabla_{a}v^{b} \equiv x^{a}\partial_{a}v^{b} + x^{a}v^{c}\Gamma_{ac}{}^{b} =0$.  
Now, we can talk about paths whose tangent vectors are parallel translated relative to themselves.  These paths are known as geodesics.  If we parameterize these curves with respect to the arc length subtended by the curve$^{2}$(which we interpret as the clock of the observer following the geodesic), then the curves satisfy the equation $\ddot x^{a} + \Gamma_{bc}{}^{a}\dot x^{b} \dot x^{c} = 0$.  If we choose the $x^{a}$ as a locally Cartesian orthonormal set of coordinates, then we can interpret $\ddot x^{a}$ as the acceleration of the $x^{a}$ position along the curve, and then interpret the other term as the acceleration of the curve.  Practically, this is a difficult game to interpret in general relativity, since time is one of the coordinates (making acceleration a subtle thing to define), and because our coordinates are completely arbitrary and need not even be (and in fact, usually are not) orthonormal.  Nonetheless, this interpretation can be useful for a few special cases--the Schwarzschild black hole metric being one of them.
Here, it's actually easiest to look at the conserved energy of a particle on the geodesic first, and then to define a force from this.  Here, in units where $G=c=1$, we have $E^{2} -1= \dot r ^{2} - \frac{2M}{r} + \left(\frac{L}{r}\right)^{2}  - \frac{2ML^{2}}{r^{3}}$, where E is the energy per unit mass of the test particle, $M$ is the mass of the central gravitating object, and $L$ is the angular momentum of the test particle.  Taking a time derivative of this, and dividing by the common factor $2\dot r$, we get:
$$0=\ddot r + \frac{M}{r^{2}}-\frac{L^{2}}{r^{3}} + \frac{3ML^{2}}{r^{4}}$$
The first three terms should be familiar from Newtonian theory--they are precisely the Newtonian gravitational force term plus the centripetal inertia of the orbit.  Relative to the clock of an observer in orbit, so long as $\frac{3ML^{2}}{r^{4}}$ is small, Newtonian motion should be indistinguishable from Relativistic motion.  So, the only difference in acceleration comes from the rotation of the orbit.  It is precisely this term that created the anamolous acceleration of Mercury that was such a puzzle to the astronomers at the turn of the century.  When Einstein observed that this term was of the right sign and magnitude to explain this anamoly was when he truly believed that he had found the correct theory of gravity.  
$^{1}$ where $\partial_{a}f$ is defined to be the derivative of $f$ with respect to the a$^{th}$ coordinate: $\partial_{a}f \equiv \frac{\partial f}{\partial x^{a}}$
$^{2}$ this sounds complicated, but is in fact what we do when measuring angles on a circle in radians--how many radians I trace out tells me how much length along the circle I've traced out since I've left the origin.  Same basic idea here.
A: Firstly, calculate the 4-velocity $\mathbf u$ of your observer, stationary or not. The 4-acceleration $\mathbf a$ is given by the covariant derivative $\nabla_{\mathbf u}\mathbf u$. To compute this, look up the formula for covariant derivative in a given coordinate system. The magnitude of acceleration felt by the observer is then $\sqrt{\mathbf a\cdot\mathbf a}$.
However there is a shortcut in the case of a stationary spacetime, meaning it has a timelike Killing vector field $\mathbf\xi$ (in some region of interest). Define the "redshift" or time-dilation factor $V:=\sqrt{-\mathbf\xi\cdot\mathbf\xi}$. This defines local stationary observers, by those whose 4-velocity is parallel to $\mathbf\xi$. Wald (1984, chapter 6, exercise 4) gives their 4-acceleration as $a^\mu=\nabla^\mu\ln V$. But the covariant derivative of a scalar is just the differential, so I think we can just write $d(\ln V)=V^{-1}dV$ -- actually this gives a dual vector. Now in Schwarzschild spacetime, let's pick the "static" Killing vector field, so the redshift factor is $V=\sqrt{1-2M/r}$. So our dual acceleration is $(1-2M/r)^{-1}M/r^2\;dr$. Now evaluate its magnitude similarly to the last equation of the first paragraph, but note here we have a dual vector so must use the inverse metric. The answer is the familiar result $(1-2M/r)^{-1/2}M/r^2$. Wald's formula makes the computation much simpler, I hope my answer didn't obscure this simplicity too much.
As for a moving observer, well it could be in a freefall orbit, in which case it won't experience any acceleration. (I have skipped over conceptual issues: what we label as "acceleration" is different between general relativity and Newtonian gravity.) In GR, you can find observer worldlines with any finite acceleration.
