Btukfyl

As far as i’m aware, gravity in General Relativity arises from the curvature of spacetime, and is equivalent to an accelerated reference frame. Objects accelerating in a gravitational field are in fact inertial and are moving through geodesics in spacetime. So it could be said then that it is not really a force, but a pseudoforce much like the coriolis effect. If so, why is it necessary to quantise gravity with a gauge boson, the graviton? And why is it necessary to unify it with the other forces?

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 it most certainly not fictitious 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. 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.

Gravity is not equivalent to an accelerated frame. It's locally equivalent to an accelerated frame. That means that a point-like observed will never be able to tell whether he/she is in a gravitational field or in an accelerating spaceship. But an observer that has some characteristic size will experience tidal forces. Tidal forces are a result of a non zero curvature of space. On the other hand going to an accelerated frame does not curve spacetime because it's just a change of variables.

The mathematical statement is that you can always find a change of coordinates which, at a given point, sets to zero the first derivatives of the metric. Namely you can make the Christoffel symbols $Gamma^mu_{nurho}$ vanish at a given point. On the other hand the second derivatives of the metric, which encode the curvature, cannot be set to zero.

There is a nice exercise that demonstrates the difference between accelerating frames and curved spacetime. Suppose you have two particles following two parallel geodesics. In curved spacetime the geodesics will not stay parallel. This is discussed in the book Schutz - A first course in general relativity at the end of Section 6.5. I'll summarize the punch line but I encourage you to check the derivation there.

Call $vec{xi}$ the vector that connects two geodesics $vec{V}$ and $vec{V}'$ initially parallel. In the book the following equation is proven
$$
nabla_Vnabla_V xi^alpha = R^alpha_{phantom{a}munubeta}V^mu V^nu xi^beta,.
$$
where $nabla_V = V^mu nabla_mu$, $nabla$ being the covariant derivative and $R^alpha_{phantom{a}munubeta}$ is the Riemann tensor, which measures the curvature. The equation above states that the variation of $vec{xi}$ along the geodesic is not zero but proportional to the curvature, thus showing that geodesics deviate from being parallel in curved space. This is a frame-independent effect and cannot be realized by going to an accelerated frame.

An observer with a characteristic size of the order of $1/sqrt{R^mu_{phantom{a}nurhobeta}}$ would be able to notice this effect.

There is no difference between waves and particles. They are identical broad concepts including any possible configuration of matter or energy.

So since gravity must be transmitted as something, whatever means through which it transmits can be called a wave or particle. Because all forces must unify, forces transmitted through quanta e.g. emag require that all other forces also be quantized.