Acceleration in general relativity

The following answer refers to proper acceleration, as measured for example by an accelerometer at a given point, it coincides with the "g-force felt by" if you wish. Also I am ignoring effects due to a non-zero size of a body, that is I am speaking about point-like particles.

About this particular sentence:

It seems like acceleration cannot really exist in spacetime: every object keeps following a geodesic.

Sadly it is not. Geodesics are exactly the type of paths of free-falling observers, namely non accelerating observers. If you look at the geodesic equation $$nabla_{dot{gamma}(tau)} dot{gamma}(tau) = 0$$ it is exactly demanding that the velocity of the path remains constant. This is however not saying that all particles follow geodesics. In principle a particle can follow any time-like curve in space-time, only the free case corresponds to geodesics.

To connect with measured acceleration you must define your observer, that means a tetrad frame at each point of the oberservers path. The observer's acceleration itself is then $$ a(tau) = nabla_{dot{gamma}(tau)} dot{gamma}(tau) = nabla_{u(tau)}u(tau),$$ where $u(tau)$ is called the 4-velocity and it is exactly the deviation from following a geodesic. (quantities above are all 4-vectors). If this observer wants to measure something in his frame, all it has to do is project said quantity into his world-line, so for the acceleration $a_p$, of some other word-line corresponding to some particle he would measure a 4-acceleration of $$ucdot a_p = g(u,a_p) = g_{munu},u^mu(tau) a_p^nu(tau)$$

Notice that here the metric $g$ is assumed to be known. This is the object that has to solve Einstein field equations (homogeneous or not and any back-reaction on the geometry is ignored). So the metric, includes any effects from any energy-momentum tensor you might want to consider and tells according to your constraints what "free-falling" means. Any deviation from that is perceived as acceleration.

Another possible case is perhaps the relative acceleration appearing when you have a parametrized family of geodesics. Here one computes the rate of change of the projection of the deviation vector of the family of geodesics on to a given geodesic, explicitly, for a family of geodesics $X^mu(tau,s)$ where $tau$ is the proper time and $s$ parametrizes the different geodesics: $$A^mu = frac{D}{dtau} (T^betanabla_beta X^mu ) $$

1)
In general relativity you extend the concept of a straight line as a path that parallel transports its own tangent vector. If we define the tangent vector to a path $x^mu (lambda)$ as $dx^mu / dlambda$, the condition it is parallel transported is
$(D / dlambda) (dx^mu / dlambda) = (dx^nu / dlambda) nabla_nu (dx^mu / dlambda) = 0$
where:
$D / dlambda$ directional covariant derivative
$nabla_mu$ covariant derivative
$lambda$ affine parameter

If $x^mu (lambda)$ is a timelike path, we can define $lambda$ as the proper time $tau$ and the tangent vector as the four-velocity $U^mu = dx^mu / dtau$. Therefore the directional covariant derivative is the four-acceleration $A^mu = (D / dtau) U^mu$.

As per definition of geodesic the four-accelation is zero. A free-falling object (geodesic) does not measure any acceleration.

2)
Instead an object at rest in a curved manifold, i.e. deviating from a geodesic, experiences an acceleration. For instance, an object on the surface of the earth measures the deviation from a geodesic as the Newtonian force of gravity.

3)
Maybe the question refers to the deviation between two geodesics. In a curved manifold two nearby geodesics accelerate from each other and the acceleration $A^mu$ between them is
$A^mu = (D^2 / dlambda^2) S^mu = R^mu_{nu rho sigma} T^nu T^rho S^sigma$
where:
$S^mu$ deviation vector
$T^mu$ tangent vector
$R^mu_{nu rho sigma}$ Riemann tensor

Note: in a curved manifold you may have both geodesic and non-geodesic trajectories. In the latter you have for instance an object at rest or a spacecraft boosted by its engines.

The main issue here is a very unclear statement of the question. I will quote it and reply.

Let's say that, from my point of view, another observer is accelerating.

ok; this means that you have defined some measure of distance and time, and by this measure adopted by you, the location of the other observer has a non-zero second-rate of change.

Now, from his point of view, he is standing still:

Yes he can always adopt that way of considering his own situation if he likes.

all he feels is an overall fictitious force of gravity, which is just a consequence of the the spacetime curvature at that point (caused by stress-energy tensor at the same point).

This is totally confused. I think you have in mind an observer subject to no force other than gravity (e.g. no electromagnetic force). In this case he does not feel a force of gravity: he does not feel any force at all. The only way I can make sense of your statement is as a muddled way of saying that the acceleration which you say the other observer has might be attributed by you to a force of gravity on him.

Now, if we accept this geometric view, It seems like acceleration cannot really exist in spacetime: every object keeps following a geodesic. When some external observer thinks that an object is accelerating, he should say instead that it is following a "curved geodesic".

The word "acceleration", like the word "velocity" can only be defined as a relative concept. Someone can have non-zero acceleration with respect to one thing and zero acceleration with respect to another. I think what you are trying to say is that a bunch of things close to one another, all in free fall, will have zero acceleration relative to each other. (Here the term "free fall" means "no force is acting except gravity"). This is the sense in which there is "no acceleration" under gravity; it means nearby geodesics do not rapidly curve away or towards one another. Therefore the different parts of a small body all fall in unison without trying to approach or recede from one another; it follows that no internal forces are needed to keep them at the same separations from one another. But the distance from one object to another some distance away can have a second derivative: think of two rocks dropped from a tall building, for example. The distance between the rocks is constant, but the distance from the rocks to the ground changes with non-zero acceleration.

To understand general relativity, I advise that you keep referring back to everyday observations such as this one in order to keep track of what the geometric statements mean.

Now, from his point of view, he is standing still: all he feels is an overall fictitious force of gravity, which is just a consequence of the the spacetime curvature at that point (caused by stress-energy tensor at the same point).

A body is following a geodesic trajectory exactly if that gravitational fictitious force is not present. We are not following geodesics in our everyday life on the surface of Earth for example.

There is a fictitious force of gravity in an accelerated frame, as a rocket keeping an artificial $g$ by letting the engines working (while there is fuel). So, it is not following a geodesic.

By the way, there is no space-time curvature in that rocket case. The Riemann tensor is zero for that frame.

Even when there is space-time curvature (the space around Earth where satelites orbit for example), the stress energy tensor is zero. It is not zero inside the planet.