I've only skimmed the Wikipedia article you link to. From a quick look I'd say the paragraphs you quote are making points about what a theory of gravity needs to look like. For example you say "Curvature of spacetime in only required in order to explain tidal forces", but what that really means is that it's impossible to have a theory of gravity without curvature. That's because any theory of gravity inevitably has to describe tidal forces. You go on to say "as long as you ignore tidal forces, you can explain gravity without curvature", but you can't ignore tidal forces so you can't explain gravity without curvature.
To take your two specific questions:
Question 1. Gravity i.e. General Relativity isn't a theory of forces: it's a theory of curvature. By focussing on the "fictitious forces" you're getting the wrong idea of how GR works. When you solve the Einstein equation you get the geometry (curvature) of space. This predicts the path a freely falling object will take. We call this a geodesic and it's effectively a straight line in a curved spacetime. If you want the object to deviate away from a geodesic then you must apply a force - and there's nothing fictitious about it.
For example, GR predicts that spacetime is curved at the surface of the Earth, and if you and I were to follow geodesics we'd plummet to the core. That we don't do so is evidence that a force is pushing us away from the geodesic, and obviously that's the force between us and the Earth. But, and it's important to be clear about this, the force is not the force of gravity, it's the force between the atoms in us and the atoms in the Earth resisting the free falling motion along a geodesic.
Question 2. Again this is really just terminology. When you're free falling "gravity" is not eliminated. Remember that "gravity" is curvature, and in fact the curvature is the same for all observers regardless of their motion. That's because the curvature tensor is the same in all co-ordinate frames. The existance of tidal forces is proof that gravity/curvature is present.
When you're free falling you are moving along a geodesic. It is true to say that there are no forces acting, but this is always the case when you are moving along a geodesic. Remember a geodesic is a straight line and objects move in a straight line when no forces are acting. There would only be a force if you deviated from the geodesic e.g. by firing a rocket motor.
Response to fiftyeight's comment: this got a bit long to put in a comment so I thought I'd append it to my original answer.
I'm guessing your thinking that if you accelerate a spaceship it changes speed, so when you stop something has happened, but when the Earth accelerates you nothing seems to happen. The Earth can apply a force to your for as long as you want, and you never seem to go anywhere or change speed. Is that a fair interpretation of your comment?
If so, it's because of how you're looking at the situation. Suppose you and I start on the surface of the Earth, but you happen to be above a very deep mine shaft (and in a vacuum so there's no air resistance - hey, it's only a thought experiment :-). You feel no force because you're freely falling along a geodesic (into the Earth), while I feel a force between me and the Earth. From your point of view the force between me and the Earth is indeed accelerating me (at 9.81ms$^{-2}$). If you measure the distance between us you'll find I am accelerating away from you, which is exactly what you'd expect to see when a force is acting. If the force stopped, maybe because I stepping into mineshaft as well, then the acceleration between us would stop, though we'd now be moving at different velocities. This is exactly what you see when you stop accelerating the spaceship.
It's true that a third person standing alongside me doesn't think I'm accelerating anywhere, but that's because they are accelerating at the same rate. It's as though, to use my example of a spaceship, you attach a camera to the spaceship, then decide the rocket motor isn't doing anything because the spaceship doesn't accelerate away from the camera.
No, we should not say that Christoffel symbols are gravity. The big reason, which really should be enough, is that they are coordinate dependent. One of the main tenets of General Relativity is that coordinates don't matter. Everything physical must be expressible in a coordinate independent and/or tensorial manner.
As I said in the comments, personally I think it's a bit ridiculous to suggest that using polar coordinates somehow brings gravity into the mix, while using Cartesian coordinates does not. The equation for a straight line changes, but you can verify using any number of methods that it's still a straight line. If polar coordinates show gravity, then where is that gravity coming from? What physical object is generating it? There was none in Cartesian coordinates.
But let me address your three points:
It is not true in absolute generality that the Christoffel symbols correspond to the gravitational field, for the reasons I gave above. A gravitational field manifests itself in the Christoffel symbols, but not the other way around. Also remember that even in Newtonian gravity the equivalence principle holds, and one might argue that only tidal forces are measurable for someone in free fall, so that's one argument in favor of the curvature.
Again, even in flat spacetime there are curved coordinates. "Special relativistic" means that the metric is $\eta_{\mu\nu}$ when expressed in locally Lorentzian (i.e. Cartesian) coordinates, not in any coordinates.
This is basically the same as 2, but see the following paragraph.
I think the deep issue is that you are misunderstanding gravity for coordinate acceleration. You actually make a very good point in your number 3 argument, but you draw the wrong conclusion. The lesson of the equivalence principle is not that acceleration is relative and hence gravity is relative. You could take it as the conclusion, but then the word "gravity" is not very useful anymore because it is coordinate-dependent.
Instead, the lesson you should take away is that "gravity" should refer to something that has a physical existence independently of the observer, and that something is tidal forces, precisely because of the equivalence principle. Since coordinate acceleration is relative, the smart thing to do is to make "gravity" mean something that is not relative.
I insist with a very important point: this is not just a matter of definition; physical reality has my back here. I say this because it turns out that every time there are tidal forces, one can identify some physical object (planet, star, whatever) responsible for it. However, sometimes objects seem to not obey the Cartestian-flat-space geodesic equation with no apparent source of gravity nearby. To me it just makes much more sense to say that the thing that always manifests itself near a heavy object is gravity and that the thing that sometimes happens as a result of weird coordinates is not gravity.
Best Answer
While it's common to describe gravity as a fictitious force we should be cautious about the use of the adjective fictitious as this is a technical term meaning the gravitational force is not fundamental but is the result of an underlying property. The force itself most certainly exists as anyone who has been sat on by an elephant can attest.
There is a sense in which all forces are fictitious since they are all the emergent long range behaviour of quantum fields, so gravity is not unique in this respect. For more on this see Can all fundamental forces be fictitious forces?
Anyhow, the object responsible for the gravitational force is a tensor field called the metric, and when we quantise gravity we are quantising the metric not the force. The graviton then emerges as the excitation of the quantum field that describes the metric. As with other quantum fields we can have real gravitons that are the building blocks of gravitational waves and virtual gravitons used in scattering calculations.
Finally, you ask why it's necessary to quantise gravity, and this turns out to be a complicated question and one that ignites much debate about what it means to quantise gravity. However the question has already been thoroughly discussed in Is the quantization of gravity necessary for a quantum theory of gravity? While it's not directly related I can also recommend A list of inconveniences between quantum mechanics and (general) relativity? as interesting reading.
The principle reason that we want to quantise gravity is because Einstein's equation relates the curvature to the matter/energy distribution and the matter/energy is quantised. Einstein's equation tells us:
$$ \mathbf G = 8 \pi \mathbf T $$
where $\mathbf G$ is the Einstein tensor that describes the spacetime curvature while $\mathbf T$ is the stress-energy tensor that describes the matter/energy distribution. The problem is that $\mathbf T$ could describe matter that is in a superposition of states or an entangled state, and that implies that the curvature must also be in a superposition of states or entangled. And this is only possible if the spacetime curvature is described by a quantum theory, or some theory whose low energy limit is quantum mechanics.