The crystal lattice symmetry imposes -- when no defect is present -- the wave function to be periodic with the unit cell length scale. What about the end of the system then ? Well, we suppose as a first try that the system is so large that the boundary are of no importance, then closing the states in the bulk as you proposed is not so stupid, since you find some solutions, and you can then compare them with experiments with more or less success.
The boundary are nevertheless important for some particular cases, especially when you have a gapped system (see Surface state on wikipedia for instance). This topic is pretty broad and really complicated to appreciate in a first year lecture on condensed matter.
So the model you're working on is good if you want to learn basics properties of matter.
Your last question has a definite "no" answer. If you allow a boundary condition like $\Psi(x+L) = 2\Psi(x)$ ($L$ being the lattice constant I presume in your head), then you will never find a normalizable wave function in an infinite space... that's bad, isn't it ?
You cannot have a total vorticity with periodic boundary conditions, since if you take a path around all of your vortices, it will have a non-zero circulation. But you have periodic bc, so you can continuously deform that path to a point, and a point has zero circulation.
Mirror images are not quite the same as in electrostatics. We want periodic boundary conditions. To get periodic boundary conditions you imagine tiling the plane with your system. This will trivially be periodic and so you can just take a tile, and work with that. So you want the mirror image of a vortex to be a vortex.
(In electrostatics you usually use mirror images to enforce not periodic boundary conditions, but to enforce constant voltage. That's why you flip the sign of the mirror charge. Here we want periodic b.c. If you flipped the sign of the vortices I believe you would get *anti*periodic boundary conditions, but don't quote me on that.)
This evolving in imaginary time is presumably an "annealing" type of operation. You are free to run the GP equation on any initial condition you want. However, to cleanly see the interaction of the vortices, we want to the vortices to be in their ground state. Otherwise when we turn on the time evolution they will get rid of their excess energy by shedding waves and other junk.
One way to get to the ground state is to evolve you equation in "imaginary time". Your usual time evolution is $\exp(i\hat{H}t)$. If plug in $t = i\tau$ you get $\exp(-\hat{H}\tau)$. Applying this to a state exponentially suppresses the higher-energy components, so you get rid of the high energy stuff. This is related to finite-temperature (just replace $\tau$ with $\beta$ and you have the partition function), but for your purposes you can just consider it a convenient mathematical trick.
Note that since you're annealing anyway, the specific details of what state you start might not be so important, since you will end in the same place anyway (hopefully).
Finally, they are a little thin on the details, so if you plan to use this work, you should just send the authors an email asking for details.
Best Answer
In a real-life misshapen blob of metal, strictly speaking the cyclic boundary conditions cannot be applied, since the blob only has a trivial group of spatial symmetries. However, the blob is approximately invariant under lattice translations (with the only mismatch occurring with the extremely small number of atoms at the surface) so it is tempting to associate the much larger symmetry group of lattice translation operators with the system.
Doing so, one obtains a system which is considerably more mathematically convenient than the otherwise rigorously-correct picture of a misshapen blob of atoms. As such, the Born-von Karman boundary conditions can be considered an approximate model to the behavior of real materials. That said, it's a pretty good approximation for macroscopically-sized materials.
That said, I don't have an explicit proof of why this approximation doesn't cause the math to explode and give wrong answers, so someone better versed than me could probably elaborate on this.