Partial answer for the first part of your question:
It is written :
"As each term commutes with the reflection operator, the full Hamiltonian must commute with the reflection operator, and thus, the eigenstates of H in the symmetry-adapted basis are either symmetric or anti- symmetric about the line defect."
"Antisymmetric states have a node at the line defect, and as a result, there are no matrix elements within the nearest-neighbor model coupling the left and right sides."
Relative to the reflection operation, then let $|f_+\rangle$ be the symmetric eigenstate and $|f_-\rangle$ be the anti-symmetric eigenstate (they are also eigenstates of the graphene hamiltonian).
So, $\langle \vec r|f_-\rangle_{|{line-defect}} = \langle \vec r|f_-\rangle_{|{\vec r= \vec 0}}= f_-(\vec 0)=0$
This is because $f_-(\vec r) = - f_-(- \vec r)$, which expresses the anti-symmetric character of the reflection eigenstate.