Coupling fission and fusion and disappear all mass. Why doesn't this happen? I'm confused about the fact that in a fission process the masses of the "products" are less than the mass of the "reactants". And in a fusion process the masses of the "products" are less than the masses of the "reactants".
My first question is then: Can I have an atom, fission it, then fusion it, then fission it, etc, etc,. And arrive at no mass and pure energy? I know this is wrong, but I don't know why so.
If you tell me this process is possible then how is it possible to recover a certain isotope of Uranium. I see this as a problem because having less mass in the "products" than in the "reactants" shows that nuclear processes aren't reversible, unless I inject mass in some way.
My second question is: Does the mass difference manifest itself as, e.g., "nucleons weigh less now", or as in "$x$ number of nucleons have disappeared".
 A: It is not true that in all fusion and fission processes the mass of the products is less than the mass of the reactants. This is only valid for exothermic reactions. 
The change of mass is due to the change in the binding energy of the nucleons (note that the change in binding energy is in the order of 1 MeV, while the mass of the nucleons is around 940 MeV). The total mass of the nucleus is the sum of the masses of its nucleons minus the binding energy, so a tighter binding means a smaller mass. The binding energy of a nucleus can be computed with Weizsäcker's semi-empirical mass formula:

In this plot, you can see the binding energy per nucleon. For isotopes lighter than iron, nuclei can become more stable if they fuse (they move to the right in the plot), and they liberate energy (lower their mass) in the process. In the other hand, fission occurs in heavier nuclei (they go to the left), and also liberates energy. But in both cases the maximum energy liberated is given by the above plot: you can only get in total, no matter what intermediate steps you make, is the 9 MeV/nucleon difference between ${}^{56}$Fe and ${}^1$H, wich corresponds to fuse 26 protons and 30 neutrons together to form an iron nucleus. And obviouslly, iron nucleii do have mass!
A: 
My first question is then: Can I have an atom, fission it, then fusion it, then fission it, etc, etc,. And arrive at no mass and pure energy? I know this is wrong, but I don't know why so.

No , you cannot. In addition to special relativity where invariant masses of complex objects are the "length" of their energy momentum summed four vector, which allows for the gain in energy, the atomic/nuclear level is quantum mechanical, not classical. 
In quantum mechanics we have in addition to energy momentum and angular momentum conservation,  conservation of quantum numbers, and in this particular proposal, it is conservation of baryon number that will not allow the scenario.
In the best/worst case,  an atomic number A nucleus may have A nucleons ( neutrons/protons) free.
In addition, in fission some of the fragments are stable , again due to quantum mechanical rules: how the energy levels are arranged in the effective nuclear potential.
In fusion , again baryon numbers are conserved and again not all states are allowed to be bound due to the nuclear potential.

My second question is: Does the mass difference manifest itself as, e.g., "nucleons weigh less now", or as in "x number of nucleons have disappeared".

The  excess energy appears as kinetic energy of the fragments and also might appear as a gamma ray ( photon) if spin quantum numbers allow it. Again, no disappearing nucleons, baryon number is conserved.
A: If uranium undergoes fission, its products are too massive to release energy in fusion. Atoms must be smaller than iron to release energy through fusion.
