What gives us the equations of motion in GR? Maybe stupid question, but to my understanding, the Einstein equation tells us the differential equation governing the Geometry of spacetime. That's all good and fine , but suppose I had an actual particle with some forces acting on it, how would I find it's trajectory? Would Newton's second law still hold?

Absurd example: Hypothetically suppose that I wanted to study block on a ramp  with friction but with the system near a black hole. If I were to do this, then how exactly can I fit in the fact that the block is effected by friction into the idea that the block follows Geodesics in space time..?
 A: When there is only gravity (i.e. no other forces),
then the equation of motion for a particle is given by the geodesic equation.
$$\frac{d^2x^\mu}{ds^2}
 +\Gamma^\mu{}_{\alpha\beta}\frac{dx^\alpha}{ds}\frac{dx^\beta}{ds}=0$$
When there is also an electromagnetic field (given by the electromagnetic
field tensor $F^\mu{}_\nu$),
then the equation of motion for a particle (with mass $m$ and charge $q$)
gets a Lorentz force on the right side.
$$m\left(\frac{d^2x^\mu}{ds^2}
 +\Gamma^\mu{}_{\alpha\beta}\frac{dx^\alpha}{ds}\frac{dx^\beta}{ds}\right)
 =q F^\mu{}_\nu \frac{dx^\nu}{ds}$$
If you want to include another force (like friction, air-drag etc.),
then you need to formulate this force as a covariant 4-vector
(possibly depending on position $x^\nu$ and velocity
$\frac{dx^\nu}{ds}$) and add it to the right side as well.
A: The answer comes in two parts.
First one may reason from the weak equivalence principle that the equation of motion of a particle in freefall must be the geodesic equation, and furthermore one may reason from the strong equivalence principle that deviations from the geodesic equation must be those given by the physics of flat spacetime in each local region, unless the physical effects in question act on scales of the order of the local radii of curvature. Thus in this first approach one holds that the Einstein equation describes spacetime, and then the equivalence principle is sufficient to show what the equation of motion of particles must be.
The result is that, yes, Newton's second law does hold (in the version ${\bf f} = d{\bf p}/dt$ with ${\bf p} = \gamma m {\bf v}$), but one has to beware of applying it correctly, because the space in which the law is being applied does not necessarily have a Euclidean geometry. In practice this means the 3-vector method has to be abandoned and one turns to 4-vectors and connection coefficients $\Gamma^a_{bc}$.
But there is a nice further feature of general relativity, namely that the field equation itself implies what the equation of motion of particles in freefall must be. This is because each small body (let's avoid considering a truly point particle so that we can have a finite mass without an infinite curvature) has, associated with it, a change in the local spacetime, a "bump in space" if you like, and the field equation tells this bump how to move!
The situation has an analogy in electromagnetism. In electromagnetism you can start with the Maxwell equations (i.e. the field equations) and then you will find your theory is incomplete, because you don't yet know how the field tells the matter how to move. So then you might bring in the Lorentz force equation. But instead of doing that, you could simply assert conservation of energy and momentum overall. Then the Lorentz force equation can be derived. In other words, in classical electromagnetism the combination of field equations and energy-momentum conservation is sufficient to derive the equation of motion of charged particles.
The similar result in General Relativity is that the combination of field equations and an assertion closely related to energy-momentum conservation is sufficient to allow one to derive the equation of motion of massive particles in freefall. But now the idea of energy-momentum conservation is itself expressed within the field equations! They do both jobs at once! To be precise, the Einstein field equation is constructed in such a way that
$$
\nabla_\lambda G^\lambda_b = 0
$$
(that is, covariant divergence of Einstein tensor is zero), and this implies
$$
\nabla_\lambda T^\lambda_b = 0,
$$
that is, covariant divergence of stress-energy tensor is zero. The latter result would express conservation of energy and momentum if spacetime were flat; in curved spacetime it is sufficient to deduce how matter moves in freefall. It is, in this sense, the equation of motion.
In the above I mentioned the covariant derivative, which is a mathematical method which will only be accessible to experts. But I hope there is some content here for non-experts too. The message is two-fold. First, in General Relativity physical effects in each small region of spacetime are predicted to go exactly as Special Relativity would say. Secondly, the equation that relates spacetime curvature to matter content also manages to express how the matter moves in response to gravity. This is the sort of thing which both fills experts with a sense of awe at the beauty of the whole structure, while also giving a sort of bafflement at how the equations manage to do two things at once.
A: Yes, Newton's first law sort of holds. You have to generalize it to curved geometries. You can think of Newton's first law as the statement that "In the absence of external forces, a particle's velocity gets parallel transported along its trajectory".
In the absence of curvature, the parallel transport is the trivial one : A straight line trajectory (assuming a nice co-ordinate system).
In the presence of curvature, there is no nice co-ordinate system which will give you the trivial straight line parallel transport. Instead, you have to use the geodesic equation, which parallel transport a vector in a general geometry.
A: The equations of motion of any matter, as well as Einstein's equation for gravitation are encoded in the Einstein-Hilbert Lagrangian :
$$
S=\int d^4x \sqrt{-g}\left[\frac{1}{16\pi G}R+\mathcal{L}_{\mathcal{M}} \right]
$$
See here for a bit more info.
The matter content of the theory (i.e., the particle you talk about for instance) will be contained in the Lagrangian $\mathcal{L}_{\mathcal{M}}$. For a massive particle for example (which trajectory is parametrised by $\tilde{x}^\mu(\lambda)$), the action would be :
$$
S_{partic}=-m\int d\lambda \sqrt{\frac{d\tilde{x}^\mu}{d\lambda}\frac{d\tilde{x}^\nu}{d\lambda}g_{\mu\nu}(\tilde{x}(\lambda))}
$$
Let us consider the simplest case in which the particle is affinely parametrised. This simplifies the action, and we can consider instead :
$$
S_{partic}=-\frac{m}{2}\int d\lambda \frac{d\tilde{x}^\mu}{d\lambda}\frac{d\tilde{x}^\nu}{d\lambda}g_{\mu\nu}(\tilde{x}(\lambda))
$$
For a single particle, it is not very convenient to use the field theory notation, so we will not extract $\mathcal{L}_{\mathcal{M}}$. Instead, the full action of the system becomes :
$$
S=\frac{1}{16\pi G}\int d^4x \sqrt{-g}R+S_{partic}
$$
The full action now depends on both the metric field $g_{\mu\nu}(x)$, as well as the particle trajectory $\tilde{x}^\mu(\lambda)$. To obtain the equations of motion, we should vary both w.r.t. to $g_{\mu\nu}$ and $\tilde{x}^\mu$. Doing the latter variation, the ricci scalar $R$ will not contribute as it does not depend on $\tilde{x}^\mu$.
So we obtain the equation of motion for a particle moving in a metric space $g_{\mu\nu}$, which is simply the geodesic equation (here I write it assuming $\lambda$ is an affine parameter) :
$$
\frac{d^2 \tilde{x}^\mu}{d^2\lambda}+\Gamma^\mu_{\rho\nu}\frac{d\tilde{x}^\rho}{d\lambda}\frac{d\tilde{x}^\nu}{d\lambda}=0
$$
Where $\Gamma^\mu_{\nu\rho}$ are the Christoffel symbols for the metric.
Now, we must do the variation w.r.t. $g_{\mu\nu}(x)$, the metric. Under this variation, we have :
$$
\delta(\sqrt{-g}R)=\sqrt{-g}(R_{\mu\nu}-\frac{R}{2}g_{\mu\nu})\delta g^{\mu\nu}
$$
It remains to compute the variation of $S_{partic}$. We will have to write $g_{\mu\nu}(\tilde{x}(\lambda))=\int d^4x g_{\mu\nu}(x)\delta^4(\tilde{x}-x)$. This is because we want to make a variation w.r.t. $g_{\mu\nu}(x)$, so we should express it as a function of $x$.
To be clearer, this is like when we say $\frac{\delta f(a)}{\delta f(x)}=\delta(x-a)$. The precise steps are $f(a)=\int dx f(x) \delta(x-a)$. Then $\delta f(a)=\int dx \delta f(x) \delta(x-a)\equiv \int \frac{\delta f(a)}{\delta f(x)}\delta f(x)$, which gives the aforementioned result.
Anyway, after rewriting the metric, and using as $\delta g_{\mu\nu}=-g_{\mu\rho}g_{\lambda\nu}\delta g^{\rho\nu}$ we find :
\begin{align}
\delta S_{partic}&=\frac{m}{2}\int d\lambda\frac{d\tilde{x}^\mu}{d\lambda}\frac{d\tilde{x}^\nu}{d\lambda}\int d^4x \delta g_{\mu\nu}\delta^4(x-\tilde{x})\\
&=-\frac{m}{2}\int d\lambda \int d^4x \left(\frac{d\tilde{x}_\rho}{d\lambda}\frac{d\tilde{x}_\nu}{d\lambda}\right)\delta g^{\nu\rho} \delta^4(x-\tilde{x})
\end{align}
Putting everything together, we obtain the following equations :
$$
 R_{\mu\nu}-\frac{R}{2}g_{\mu\nu}=16\pi G\left(\frac{m}{2}\int d\lambda \frac{d\tilde{x}_\nu}{d\lambda}\frac{d\tilde{x}_\mu}{d\lambda} \delta^4\left(x-\tilde{x}(\lambda)\right)\right)
$$
On the RHS, this is simply the Stress-energy tensor of a single particle. All in all, we end up with two differential equations which are coupled. Solving these coupled equations will give you both the particle trajectory, and the metric of spacetime, which will include the backreaction of the particle on the spacetime as it traverses it.
However, these coupled equations are very difficult to solve. What one does in practice is assume the backreation of the particle on the spacetime is small. This will have for effect to decouple the system of equation. Then, what you would do is first solve the Einstein equations to find the metric, and THEN solve the particle geodesic equations to find the particle trajectory. In that approach, you neglect how the particle will deform the spacetime as it moves.
