Let me try to approach this description purely from a quantum mechanical (and therefore a QFT point of view) for gravity. Newton's potential is seen as the tree-level Feynman diagram in quantum gravity. The tree level Feynman diagram can be calculated as
\begin{equation}
i\mathcal{M}(q) = \tau_{\rho\sigma}(k_1,k_2;m_1)\frac{i\mathcal{P}^{\rho\sigma\alpha\beta}}{q^2}\tau_{\alpha\beta}(k_3,k_4;m_2)
\end{equation}
where $\tau_{\mu\nu}$ is the 3-point vertex for 2 scalar and 1 graviton interaction and
\begin{equation}
\mathcal{P}^{\rho\sigma\alpha\beta} = \frac{1}{2}\left(\eta^{\rho\alpha}\eta^{\sigma\beta} + \eta^{\rho\beta}\eta^{\sigma\alpha} - \eta^{\rho\sigma}\eta^{\alpha\beta}\right)
\end{equation}
I suppose I can also write down the 3-vertex which is (if I can recall from memory)
\begin{equation}
\tau^{\mu\nu}(k_1,k_2;m) = -\frac{i\kappa}{2}\left(k_1^\mu k_2^\nu + k_1^\nu k_2^\mu - \eta^{\mu\nu}(k_1\cdot k_2 - m^2)\right).
\end{equation}
After using these Feynman rules and performing contractions you get a large mess. The way to reduce it is to consider a particular time-slice in the Feynman diagram such that $k_1\cdot k_2 = m_1^2$, $k_1\cdot k_3 = m_1 m_2$ and $q^2 = -\mathbf{q}^2$ (and similarly for the rest). Once you do this you should find something that looks like
\begin{equation}
i\mathcal{M}(\mathbf{q}) = \frac{im_1m_2\kappa^2}{2\mathbf{q}^2}
\end{equation}
(modulo a minus sign I may have forgotten). I should let you know that here $\kappa^2 = 32\pi G$ where $G$ is Newton's constant. The way to Fourier transform this matrix element (which is the full amplitude at tree level since we didn't have to consider any information about the helicity) is
\begin{equation}
V(r) = \frac{1}{2m_1}\frac{1}{2m_2}\int\frac{d^3q}{(2\pi)^3}\mathcal{M}(\mathbf{q})e^{i\mathbf{q}\cdot\mathbf{r}}.
\end{equation}
At the end of this long calculation, you see that you started with a quantum mechanical theory, and after a few assumption, namely one that reduces you to the IR limit of the theory, we recover Newton's potential. Now, with this exact same formalism, you can derive quantum mechanical corrections to Newton's potential, namely of the form
\begin{equation}
V(r) = -\frac{Gm_1m_2}{r} + a\frac{G^2\hbar}{r^2}
\end{equation}
where $a$ depends upon what matter you put into the 1-loop correction. It turns out that in pure gravity, $a = 41/10\pi$. This factor changes when you include more and more matter. So, the loop-correction is a seriously hard-calculated quantum mechanical calculation and prediction, but we also derived Newton's potential from this manner as well.
Now, this was all QFT, but we can do something similar with basic quantum mechanics, namely to include Newton's potential into your Schrodinger equation
\begin{equation}
i\hbar\frac{\partial\psi}{\partial t} = -\frac{\hbar^2}{2m}\nabla^2\psi + V\psi + m\phi\psi
\end{equation}
where $V$ is some ordinary potential and $\phi$ is the gravitational potential. The solution leads to the Schrodinger-Newtons equation. You can derive that the stationary ground state has a width of $a_0\sim \hbar^2/Gm^3$.
Hopefully this answers your overall question(s).
