How are pi-orbitals delocalized in a porphyrin chain?
What you are seeing are molecular orbitals which, by definition, extend across the entire molecule, restricted only by inherent symmetry. It is important to note and is not clearly visible in your image that you are looking at a $\ce{H2-Fe-Zn-H2}$ system, i.e. from one side to the other the centres of the porphyrin system contain two hydrogens, an iron ion, a zinc ion and another two hydrogens. Replace the zinc with iron or vice-versa and you would get an axis of symmetry with molecular orbitals that are either perfectly symmetric or perfectly antisymmetric to this axis of symmetry.

As it is drawn, the molecule is $C_\sigma$, i.e. there is a single plain of symmetry along the plain of the molecule. All orbitals must be symmetric or antisymmetric to this plain, and since we are looking at a $\unicode[Times]{x3C0}$-system, it will be antisymmetric.

All the rest is the result of a calculation! So we cannot really answer why, all we can say is ‘because that is what maths gives us’. But we can slightly rationalise it. First, there is a difference between the top and the bottom, it is just not as easily visible. This is because the only difference between the top and the bottom is ‘where exactly are the two hydrogens and the double bonds?’ Second, the two ions are clearly different ones with different electronegativities that create entirely different environments. Therefore, it is understandable that one orbital is centred more around one of the two ions and the other around the other. And third, the LUMO$+1$ is clearly delocalised across the entire molecule.