How do you derive Newtonian Sum of Forces equation from GR? This is what I have so far.
We start with the definition that force is the change in momentum with respect to time:$$\Sigma F_i =\frac{dP}{d\tau}$$Where $\Sigma F_i$ is the sum of forces, both applied and inertial, $P$ is the momentum and $\tau$ is the proper time. For a constant mass system (non-relativistic), this is:$$\Sigma F_i =m\frac{dU}{d\tau}$$Where $U$ is the velocity.  In spacetime, the velocity is a four-velocity and is the product of two tensors, the components and the basis vectors.  Expanding, we get:$$\frac{dU}{d\tau}=\frac{d}{d\tau}\left(U^\lambda e_\lambda \right)=\frac{dU^\lambda}{d\tau}e_\lambda +U^\lambda\frac{de_\lambda}{d\tau}$$
The derivative of the basis vector with respect to the affine parameter is represented by the Christoffel Symbols, so the product of the tangent velocity of the fiduciary geodesic and the rate of change in the basis vector is (please provide the derivation if you consider it important):$$U^\lambda\frac{de_\lambda}{d\tau}=\Gamma_{\mu\nu}^\lambda U^\mu U^\nu e_\lambda$$
Substituting back in, the expression for the four-acceleration becomes:$$\frac{dU}{d\tau}=\frac{dU^\lambda}{d\tau}e_\lambda +\Gamma_{\mu\nu}^\lambda U^\mu U^\nu e_\lambda$$$$\frac{dU}{d\tau}=\left(\frac{dU^\lambda}{d\tau}+\Gamma_{\mu\nu}^\lambda U^\mu U^\nu\right)e_\lambda$$Substituting back into the sum of forces equation, we have:$$\Sigma F_i =m\left(\frac{dU^\lambda}{d\tau}+\Gamma_{\mu\nu}^\lambda U^\mu U^\nu\right)e_\lambda$$If the geometry of spacetime is flat, then the Christoffel symbols disappear and, for a non-relativistic, constant mass system, we have: $$a\equiv\frac{dU}{d\tau}$$
$$\Sigma F_i =ma$$If the spacetime is curved, we have:$$A\equiv\Gamma_{\mu\nu}^\lambda U^\mu U^\nu$$$$\Sigma F_i =m\left(a+A\right)$$Where $a$ is the result the applied force and $A$ is the acceleration of the pseudo (inertial) force of curvature.
I cobbled this together from several papers.  I would appreciate any changes that make the notation and the descriptions more standard.  In particular, I'm not sure of the treatment of the basis vector, $e^\lambda$ in all the steps.
 A: If you use the "full" 4-vector (not the components) you can't put indices, so for example the equations should rather be:
$$\frac{dU}{d\tau}=\frac{d}{d\tau}\left(U^\lambda e_\lambda \right)=\frac{dU^\lambda}{d\tau}e_\lambda +U^\lambda\frac{de_\lambda}{d\tau}$$
$$\frac{dU}{d\tau}=\left(\frac{dU^\lambda}{d\tau}+\Gamma_{\mu\nu}^\lambda U^\mu U^\nu\right)e_\lambda$$
$$\Sigma F_i =m\left(\frac{dU^\lambda}{d\tau}+\Gamma_{\mu\nu}^\lambda U^\mu U^\nu\right)e_\lambda$$
Etc. etc.
A: Ok I'm adding another answer because I didn't find a textbook with a full clear derivation like the one I'm gonna write.
So, your post with my edits is perfectly correct but only in Special Relativity, I will also clarify this at the end.
In General Relativity you need Covariant Derivatives, but the final result will be the same, because we will basically swap $\partial$ with $\nabla$.
So let's start with some proprieties and definitions:

*

*Notation:
If there are indices I'm indicating the components of the 4-vector (ex. $U^{\mu}$) otherwise I'm indicating the full 4-vector (ex. $U=U^{\mu}e_{\mu}$). $\tau$ stands for proper time (in general we should use $s$, the parameter of the curve along which lives the system we are considering, for a mass system (timelike curve) $s$ coincides with $\tau$).

*A propriety of 4-Velocity: The 4-velocity is defined as: $$U^{\mu}=\frac{d x^{\mu}}{d \tau}$$ Therefore in a coordinate basis we have: $$U=U^{\mu}e_{\mu}=U^{\mu}\partial_{\mu}=\frac{d x^{\mu}}{d \tau}\frac{\partial}{\partial x^{\mu}}=\frac{d}{d\tau}$$ Where in the last passage we used the chain rule.

*Christoffel Symbols :
The results was indeed correct because we used this definition of the Christoffel Symbols: $\Gamma^{\sigma}_{\mu \nu}e_{\sigma}=\partial_{\mu}e_{\nu}$, but in GR the right definition is: $$\Gamma^{\sigma}_{\mu \nu}e_{\sigma}=\nabla_{\mu}e_{\nu}$$

*$\nabla$ Proprieties: If $f$ is a function we have $\nabla_{\nu}f=e_{\nu}(f)$, that in a coordinate basis it becomes: $$\nabla_{\nu}f=\partial_{\nu}f$$ Obviously the components of a 4-vector are functions so we can write for example: $$\nabla_{\nu}(U^{\lambda})=\partial_{\nu}U^{\lambda}$$ This has not to be confused with $\nabla_{\nu}U^{\lambda}$ because this is just a notation for $(\nabla_{\nu}U)^{\lambda}$, therefore when we write $\nabla_{\nu}U^{\lambda}$ we mean the $\lambda$ component of the covariant derivative of $U$ in the direction $e_{\nu}$; so $(\nabla_{\nu}U)^{\lambda}:=\nabla_{\nu}U^{\lambda}$.
As in the flat case, the Covariant Derivative along a vector $V=V^{\rho}e_{\rho}$ is: $$\nabla_{V}U=\nabla_{V^{\rho}e_{\rho}}U=V^{\rho}\nabla_{\rho}U$$ I said as in the flat case because if you remember the directional derivative is $\partial_V=V^{\rho}\partial_{\rho}$.

*Force: In GR the 4-acceleration is the Covariant Derivative of the 4-velocity along the direction of the 4-Velocity: $$a=\nabla_{U}U$$ Indeed we have that the geodesic equation is $\nabla_{U}U=0$, but if there is acceleration we are nomore along a geodesic so in this case $\nabla_{U}U \neq 0 =a$. Since that we are using 4-vectors, with these we can still write the "2nd Newton's law" as: $$F=m a=m\nabla_{U}U$$
Ok now we are ready to derive the full expression of the 4-force:
$$F=m \nabla_{U}U=m U^{\nu}\nabla_{\nu}U=m U^{\nu}\nabla_{\nu}(U^{\mu}e_{\mu})=m[ U^{\nu}\nabla_{\nu}(U^{\mu})e_{\mu}+U^{\nu}U^{\mu}\nabla_{\nu}(e_{\mu})]=m[ U^{\nu}(\partial_{\nu}U^{\mu})e_{\mu}+U^{\nu}U^{\mu}\Gamma^{\sigma}_{\nu \mu}e_{\sigma}]=m[ U^{\nu}(\partial_{\nu}U^{\mu})e_{\mu}+U^{\nu}U^{\sigma}\Gamma^{\mu}_{\nu \sigma}e_{\mu}]=m[ U^{\nu}\partial_{\nu}U^{\mu}+U^{\nu}U^{\sigma}\Gamma^{\mu}_{\nu \sigma}]e_{\mu}=m\left[ \frac{dU^{\mu}}{d \tau}+U^{\nu}U^{\sigma}\Gamma^{\mu}_{\nu \sigma}\right]e_{\mu}$$
Since obviously $F=F^{\mu}e_{\mu}$ we can write the equation using just the components:
$$F^{\mu}=m\left[ \frac{dU^{\mu}}{d \tau}+U^{\nu}U^{\sigma}\Gamma^{\mu}_{\nu \sigma}\right]=m \nabla_{U}U^{\mu}$$
The result was obviously immediate from the proprieties $\nabla_{V}=V^{\lambda}\nabla_{\lambda}$, the expressions for the 4-velocity $U^{\nu}\partial_{\nu}=\frac{d}{d \tau}$ and for the covariant derivative $\nabla_{\nu}V^{\mu}=\partial_{\nu}V^{\mu}+V^{\sigma}\Gamma^{\mu}_{\nu \sigma}$.
I hope that the derivation is clear, sorry but I don't know how to label equations and recall them in math mode.

*

*Conclusions
As you can see the result is exactly the same, but this has a very very strong implication, this exactly means that Gravity is just an Apparent Force.
I said that your derivation was right in SR but it didn't holds in GR. This is because the Christoffel symbols $\Gamma$ are used to describe Apparent Forces, with the definition $\Gamma^{\sigma}_{\mu \nu}e_{\sigma}=\partial_{\mu}e_{\nu}$ you are only considering the Apparent Forces due to not being in an inertial frame, like the Centripetal Force. But if the manifold is not flat we must also consider the curvature, and defining the symbols as $\Gamma^{\sigma}_{\mu \nu}e_{\sigma}=\nabla_{\mu}e_{\nu}$ does exactly this.
Why gravity then? Well, the equivalence principle tell us that an observer in free fall lies on a geodesic, while the others none; So, for example, we on the Earth are not on a geodesic while the astronauts in the ISS as free falling.  So the GR teach us that THEIR frame is an inertial frame, while OUR none. So the gravity we are observing pulling the astronauts is just an Apparent Force due to the fact that WE are not in a inertial frame, mathematically contained in the Christoffel symbols.
