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.

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).

Let $\vec r_\parallel$ and $\vec r_\perp$ be the directions parallel and orthogonal to the line defect.

So, $\langle \vec r|f_-\rangle_{|{line-defect}} = \langle \vec r|f_-\rangle_{|{\vec r_\perp= \vec 0}}= f_-(\vec r_\parallel,\vec r_\perp=\vec 0)=0$

This is because $f_-(\vec r_\parallel, \vec r_\perp) = - f_-(\vec r_\parallel, -\vec r_\perp)$, which expresses the anti-symmetric character of the reflection eigenstate.