About negative energies: they set no problem:
On this context, only energy differences have significance. Negative energy appears because when you've made the integration, you've set one point where you set your energy to 0. In this case, you have chosen that $PE_1 = 0$ for $r = \infty$. If you've set $PE_1 = 1000$ at $r = \infty$, the energy was positive for some r.
However, the minus sign is important, as it is telling you that the test particle is losing potential energy when moving to $r = 0$, this is true because it is accelerating, causing an increase in $KE$:
let's calculate the $\Delta PE_1$ for a particle moving in direction of $r = 0$: $r_i = 10$ and $r_f = 1$:
$\Delta PE_1 = PE_f - PE_i = Gm(-1 - (-0.1)) = -Gm\times0.9 < 0$
as expected: we lose $PE$ and win $KE$.
Second bullet: yes, you are right. However, it is only true IF they are point particles: has they normally have a definite radius, they collide when $r = r_1 + r_2$, causing an elastic or inelastic collision.
Third bullet: you are right with $PE_2 = mgh$, however, again, you are choosing a given referential: you are assuming $PE_2 = 0$ for $y = 0$, which, on the previous notation, means that you were setting $PE_1 = 0$ for $ r = r_{earth}$.
The most important difference now is that you are saying that an increase in h is moving farther in r (if you are higher, you are farther from the Earth center).
By making the analogy to the previous problem, imagine you want to obtain the $\Delta PE_2$. In this case, you begin at $h_i = 10$ and you want move to $h_f = 1$ (moving in direction to Earth center, like $\Delta PE_1$:
$\Delta PE_2 = PE_{f} - PE_{i} = 1mg - 10mg = -9mg < 0$.
As expected, because we are falling, we are losing $PE$ and winning $KE$, the same result has $PE_1$
Fourth bullet: they both represent the same thing. The difference is that $gh$ is the first term in the Taylor series of the expansion of $PE_1$ near $r = r_{Earth}$. As exercise, try to expand $PE_1(r)$ in a taylor series, and show that the linear term is:
$PE_1 = a + \frac{Gm(r-r_{earth})}{r_{earth}^2}$.
Them numerically calculate $Gm/r_{earth}^2$ (remember that $m=m_{earth}$). If you haven't made this already, I guess you will be surprised.
So, from what I understood, your logic is totally correct, apart from two key points:
energy is defined apart of a constant value.
in the $PE_1$, increase r means decrease $1/r$, which means increase $PE_2 = -Gm/r$. In $PE_2$, increase h means increase $PE_2=mgh$.
We introduce a minus sign to equate the mathematical concept of a potential with the physical concept of potential energy.
Take the gravitational field, for example, which we approximate as being constant near the surface of Earth. The force field can then be described by $\vec{F}(x,y,z)=-mg\hat{e_z}$, taking the up/down direction to be the $z$ direction. The mathematical potential $V$ would be $V(x,y,z) = -mgz+\text{Constant}$ and would satisfy $\nabla V=\vec{F}$. This would correspond with decreasing in height increasing in potential energy which would make us have to redefine mechanical energy as $T-V$ in order to maintain conservation.
Instead of redefining mechanical energy, we introduce the minus sign $\vec{F} = -\nabla V$ which equates the physical notion of potential energy with the mathematical notion of the scalar potential.
Best Answer
When you do conservative work on an object, the work you do is equal to the negative change in potential energy $W_c = - \Delta U$. As an example, if you lift an object against Earth's gravity, the work will be $-mgh$. Gravity is doing work on the object by pulling it towards the Earth, but since you are pushing it in the other direction, the work you do on the box (and therefore the force) is negative. The field does negative work when you increase a particle's potential energy.
Mathematically, it is just that $F=\frac{dW}{dx}$, which means that if the work is conservative, then $F=\frac{-dU}{dx}$, since $W_c = - \Delta U$. Then $-dU = Fdx$, so $U = - \int F dx$.
We can also say that work is negative when the force and displacement are in opposite directions, since $W = \vec F \cdot d\vec x = Fdxcos\phi$. When $\phi=\pi$, then $\cos\phi = -1$. An example of this conceptually is friction. An object sliding down a plane has kinetic friction acting on it. The friction is in the direction (up the ramp) opposite to the object's motion/displacement. So we say that the friction force is doing negative work.