The two quantities are on opposite sides of Newton's second law equation $\vec F=m\,\vec a$
The force on a mass $m$ in a gravitation field $\vec g (= g \,\hat d)$ is $\vec F = m \,\vec g = m\, g\,\hat d$ where $g$ is the magnitude of the gravitational field strength and $\hat d$ is the unit vector in the down direction.
Assuming no air resistance then using this force and Newton's second law you can find the acceleration of the mass in free fall.
$\vec F =m\, \vec a \Rightarrow m\, g\,\hat d = m\,\vec a = m\,a\, \hat d \Rightarrow \vec a = a \,\hat d = g \,\hat d$ where $a$ is the magnitude of the acceleration.
So the acceleration of free fall $\vec a$ has the same magnitude as the gravitational field strength $g$ and is in the same direction $\hat d$.
To differentiate between the two quantities you can use $\rm N\, kg^{-1}$ as the unit of gravitational field strength and $\rm m\, s^{-2}$ as the unit of acceleration although dimensionally they are the same.
Statements 1-4 are indeed correct1. According to Newtonian gravity, any two objects with non-zero masses impart a force on each other. Additionally, the force falls off as $r^{-2}$, so the force goes to zero only as $r\to\infty$. There's no finite cutoff. Now, there are obviously distances at which the force is too small to measure, and this is especially true for bodies with very little mass, such as atoms.
Statement 5 has me confused. You could say that an object at point $p$ a distance $r_p$ from another object with mass indeed feels the force of gravity from that other object, but I'm not sure how you would define it as "existing at both points at the same time". Gravity propagates at the speed of light (see this, this, and this), so gravitational effects aren't instantaneous, but both objects will indeed feel some sort of force.
Your final conclusion involves a bit more thinking. if an object is in a potential $\Phi(\mathbf{r})$, it will feel a force, given by the equation
$$F(\mathbf{r})=-\frac{d\Phi(\mathbf{r})}{dr}$$
where $\mathbf{r}$ is a position vector. If we can measure the force at a bunch of different $\mathbf{r}$, we can figure out $\Phi(\mathbf{r})$2. This, in turn, will allow us to figure out the density distribution of the matter causing the potential, by Gauss's law for gravity:
$$\nabla^2\Phi(\mathbf{r})=4\pi G\rho$$
In your case, it will tell us that the object being observed is being influenced by what is essentially a point mass. This approximation works because
- Atoms are extremely small and have very little mass.
- The atoms exerting the force are far away.
We're not going to get much good information about the structure of the cluster; it will seem like a point mass to us. The force between the cluster of atoms is going to be incredibly small, and this is what makes your idea not very feasible (see Has gravity ever been experimentally measured between two atoms?). Try plugging numbers into Newton's law of gravitation, taking the two masses to be the mass of a hydrogen atom and a small cluster of hydrogen atoms. Over any reasonable distances, the force of gravity between the two is going to be pretty much impossible to measure.
Another difficult arising here is that there are plenty of other effects - electromagnetic forces and other major sources of gravity, for starters - that will make any measurable force here insignificant. Using Poisson's equation is pointless (pun not intended); it would only tell us remotely useful information if used in an isolated system composed of only these atoms.
1 Statement 2 is actually a little wrong. The Sun is, largely, composed of plasma, which includes protons which have been stripped of their elections. But I think I'm being a little pedantic, as the mass of an electron is negligible compared to the mass of a proton.
If we can only measure the potential at one point, though, we don't know much about the global behavior of the density.
Best Answer
You need to keep the direction in mind. While the direction never makes the gravity negative, adding opposite directions will cancel out.
I don't know if you are familiar with vectors? The length of a vector is always positive (strength of gravity) but it also has a direction (direction of gravity). If you add two vectors of equal length ("strength") but with opposite direction they cancel out.