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.
To really understand this you should study the differential geometry of geodesics in curved spacetimes. I'll try to provide a simplified explanation.
Even objects "at rest" (in a given reference frame) are actually moving through spacetime, because spacetime is not just space, but also time: apple is "getting older" - moving through time. The "velocity" through spacetime is called a four-velocity and it is always equal to the speed of light. Spacetime in gravitation field is curved, so the time axis (in simple terms) is no longer orthogonal to the space axes. The apple moving first only in the time direction (i.e. at rest in space) starts accelerating in space thanks to the curvature (the "mixing" of the space and time axes) - the velocity in time becomes velocity in space. The acceleration happens because the time flows slower when the gravitational potential is decreasing. Apple is moving deeper into the graviational field, thus its velocity in the "time direction" is changing (as time gets slower and slower). The four-velocity is conserved (always equal to the speed of light), so the object must accelerate in space. This acceleration has the direction of decreasing gravitational gradient.
Edit - based on the comments I decided to clarify what the four-velocity is:
4-velocity is a four-vector, i.e. a vector with 4 components. The first component is the "speed through time" (how much of the coordinate time elapses per 1 unit of proper time). The remaining 3 components are the classical velocity vector (speed in the 3 spatial directions).
$$ U=\left(c\frac{dt}{d\tau},\frac{dx}{d\tau},\frac{dy}{d\tau},\frac{dz}{d\tau}\right) $$
When you observe the apple in its rest frame (the apple is at rest - zero spatial velocity), the whole 4-velocity is in the "speed through time". It is because in the rest frame the coordinate time equals the proper time, so $\frac{dt}{d\tau} = 1$.
When you observe the apple from some other reference frame, where the apple is moving at some speed, the coordinate time is no longer equal to the proper time. The time dilation causes that there is less proper time measured by the apple than the elapsed coordinate time (the time of the apple is slower than the time in the reference frame from which we are observing the apple). So in this frame, the "speed through time" of the apple is more than the speed of light ($\frac{dt}{d\tau} > 1$), but the speed through space is also increasing.
The magnitude of the 4-velocity always equals c, because it is an invariant (it does not depend on the choice of the reference frame). It is defined as:
$$ \left\|U\right\| =\sqrt[2]{c^2\left(\frac{dt}{d\tau}\right)^2-\left(\frac{dx}{d\tau}\right)^2-\left(\frac{dy}{d\tau}\right)^2-\left(\frac{dz}{d\tau}\right)^2} $$
Notice the minus signs in the expression - these come from the Minkowski metric. The components of the 4-velocity can change when you switch from one reference frame to another, but the magnitude stays unchanged (all the changes in components "cancel out" in the magnitude).
Best Answer
Suppose you and I start on the equator, a kilometre apart, and we both head exactly due North in a straight line, so we head off in exactly parallel directions:
Now we know that in Euclidean geometry parallel lines remain the same distance apart. But if you and I measure the distance, $d$, between us we find that $d$ starts off at 1km but decreases as we head North and we eventually meet at the North Pole.
Se we have a paradox: we started out parallel but we moved together. The only explanation is that there is some force pulling us together. But we know there is no force really, it's just that we are moving on a curved surface.
This is what happens in general relativity, though as I'm sure you'd expect it's a lot more complicated (principally because time is curved as well). If you see a freely falling body accelerating towards the earth you'd say there must be a force acting between the body and the Earth, and you'd call that force gravity. But the general relativist would say the Earth and the object are both moving along geodesics, i.e. in a straight line, and it's just that because spacetime is curved the two straight lines converge just as we saw for motion on a sphere. There isn't really a force acting even though it looks like a force to us. That's why gravity is sometimes described as a fictitous force.
Actually there is an accelerating object involved in this, and it's you standing on the Earth's surface. How do you know you're accelerating? Well the Earth is pushing at the soles of your shoes and accelerating you upwards. Where there's a force there's an acceleration, so the conclusion must be that the surface of the Earth is accelerating you outwards while the freely falling object you're watching is not accelerating.
If you're interested, twistor59's answer to What is the weight equation through general relativity? explains how to calculate this acceleration, though you may find the maths involved a bit hard going.