Is $\partial_\alpha f^\alpha$ coordinate-independent? At this point in Schuller's 9th lecture on GR, he claims that Poisson's equation for the Newtonian gravitational field strength is $$-\partial_\alpha f^\alpha=4\pi G \rho,$$
where $\alpha=1,2,3$. But this equation is usually written as $$\vec{\nabla} \cdot \vec{g} = -4\pi G\rho.$$ 
I can see that $\partial_\alpha f^\alpha$ and $\vec{\nabla} \cdot \vec{f}$ are identical in Cartesian coordinates, but the divergence in spherical coordinates (and others) is more certainly complicated than just $\partial_\alpha f^\alpha$. So what justifies Schuller's assertion that the above truly is Poisson's equation in a coordinate-independent setting?
 A: The divergence defined in terms of the covariant derivative,
$$\nabla_i f^i $$
is indeed coordinate independent; that's the whole point of a covariant derivative. However, as you said, the divergence is not just equal to $\partial_i\, f^i$ in general. You can relate the two either by expanding out the covariant derivative,
$$\nabla_i\, f^i = \partial_i\, f^i + \Gamma^i_{ji}\, f^j$$ 
or by the formula
$$\nabla_i\, f^i = \frac{1}{\sqrt{g}} \partial_i \left(\sqrt{g} f^i\right).$$
Schuller is probably working in a special coordinate system or class of coordinate systems where these extra terms vanish, for simplicity.
A: *

*A major point of Schuller's 9th lecture is that Newtonian spacetime is curved from a certain perspective. In fact that's the very title of the lecture. 

*At 14:15 he introduces a truncated 3-acceleration
$$ \ddot{x}^{\alpha}+\sum_{\beta, \gamma=1}^3\Gamma^{\alpha}{}_{\beta\gamma}\dot{x}^{\beta}\dot{x}^{\gamma}~=~\ddot{x}^{\alpha}, \qquad \alpha ~\in~\{1,2,3\}, $$
thereby implicitly implying that all the spatial gamma symbols $\Gamma^{\alpha}{}_{\beta\gamma}=0$ are zero. At this point he is very brief:

"This is not the acceleration, but let's not talk about it right now. That comes later."


*Just before the 46 minute mark he identifies the specific force with 3 gamma symbols $$f^{\alpha} ~=~ -\Gamma^{\alpha}{}_{00}, \qquad \alpha ~\in~\{1,2,3\},$$ and he deduces that all the other gamma symbol components should vanish in order for Newton's 2nd law to take the form of the geodesic equation in the 4-dimensional spacetime.

*So yes, the missing term $\sum_{\beta=1}^3\Gamma^{\beta}{}_{\alpha\beta}$ in the spatial 3-divergence formula 
$$\vec{\nabla} \cdot \vec{f} ~=~\sum_{\alpha=1}^3\nabla_{\alpha} f^{\alpha}~=~\sum_{\alpha=1}^3\left(\partial_{\alpha}+ \sum_{\beta=1}^3\Gamma^{\beta}{}_{\alpha\beta}\right) f^{\alpha}$$
is here assumed zero.
