What does negative potential energy mean here?
Not much. The particular value of potential energy isn't important at all in classical physics. But changes in potential energy are. You could shift everything up so that $U>0$ everywhere, and you'd still get the same physics. (Why, you ask? Well, would the force change?)
So why would people choose to have negative potential energy? Read on.
Where is the "zero" reference defined?
The typical choice one makes in systems where there is no interaction at infinite distance is to set $U=0$ at infinity. That's the zero reference here; $U(r=\infty)=0$ (in non-careful physicist speak). The fact that there is a negative potential energy is a consequence of this (and of having a long-range attractive force).
Why is there reference to repulsion and attraction in the PE graph?
From far away the molecules are attracted to each other, while at close range they repel each other. You will have to read your textbook more closely to see why your problem is choosing to model the interaction like this. But the attractive term that goes as $-1/r$ looks like Coulomb attraction to me. Others might be able to identify the repulsive $1/r^8$ term.
By the way, take a look at the Lennard-Jones potential if you're unfamiliar with it. It's a way to model the attraction and repulsion of two neutral molecules.
(These next two point are related, so don't read just one of them.)
[...] shouldn't the graph of the attractive force be flipped upside down?
They've done something a bit odd and defined an attractive force to be positive, which seems to go against the coordinate system in the potential energy diagram. An attractive force (meaning toward the origin) should have a negative sign (if $r>0$).
So it seems they're using $F=dU/dr$ for the relation between force and potential energy. This is related to the next question...
Why isn't it defined as the negative work in this system?
It depends what is meant by $F$. Most physicists would write $U=-\int F\,dr$ and know that, loosely speaking, $F$ is the conservative force between the two particles. But you could choose to have $F$ be an external force that overcomes the internal one, which would introduce another negative sign.
So perhaps they mean that $F$ is the force you would have to exert in order to keep the two atoms stationary. In the "attractive" region, they want to move toward each other, so the force on whatever particle whose position is labeled by $r$ should be negative, meaning you need to exert a positive force.