I am given 2 matrices which have the following traits:
Let $A$ and $B$ be those matrices and $a_{i,j}$ and $b_{i,j}$ be the entries of both matrices.
There are 2 disjoint subsets of the indexes, let them be defined as $G,B$.
For all $a_{i,j}$ and $b_{i,j}$ such as $i,j \in G$ or $i,j \in B$, $a_{i,j}=b_{i,j}$.
For each entry $b_{i,j}$ or $b_{j,i}$ such that $i \in G$ and $j \in B$, $-a_{i,j}=b_{i,j}$ .
Possible values in $A$: 0,1
Possible values in $B$: 0,1,-1
For example:
size of matrix = 3x3
G = {1,3}, B = {2}
$$ A = \begin{bmatrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}, \quad B = \begin{bmatrix} 0 & -1 & 1 \\ 0 & 0 & -1 \\ 1 & 0 & 0 \end{bmatrix} $$
Now, Let $W=A^TA$ and $M=B^TB$.
Certainly, $W$ and $M$ are symmetrical and $M$ may contain negative entries only in items which are with indexes from both $G$ and $B$.
What can I say about the eigenvectors and eigenvalues of $W$ and $M$?
More specifically, are the eigenvector entries which are $i \in G$ of the primary eigenvectors of $W$ and $M$ equal?
By reindexing, you may assume that $G=\{1,2,\ldots,k\}$ and hence $A$ and $B$ are of the form $$ A = \pmatrix{X&Y\\ Z&W},\quad B=\pmatrix{X&-Y\\ -Z&W}. $$ For instance, in your example, $A$ and $B$ are permutation-similar to the following partitioned matrices $\widetilde{A}$ and $\widetilde{B}$: \begin{align*} A&=\pmatrix{1&0&0\\ 0&0&1\\ 0&1&0} \underbrace{\left(\begin{array}{cc|c}0&1&1\\ 1&0&0\\ \hline0&1&0 \end{array}\right)}_{\large\widetilde{A}} \pmatrix{1&0&0\\ 0&0&1\\ 0&1&0},\\ B&=\pmatrix{1&0&0\\ 0&0&1\\ 0&1&0} \underbrace{\left(\begin{array}{cc|c}0&1&-1\\ 1&0&0\\ \hline0&-1&0 \end{array}\right)}_{\large\widetilde{B}} \pmatrix{1&0&0\\ 0&0&1\\ 0&1&0}. \end{align*} Therefore, by reindexing the rows and columns (which is in effect a similarity transform via a permutation matrix), we may assume that $A=\widetilde{A}$ and $B=\widetilde{B}$. Now, $$ \pmatrix{X&-Y\\ -Z&W} =\underbrace{\pmatrix{I_k&0\\ 0&-I_{n-k}}}_D\pmatrix{X&Y\\ Z&W}\pmatrix{I_k&0\\ 0&-I_{n-k}}. $$ That is, $B = DAD$. Since $D$ is a real orthogonal matrix, $B$ is orthogonally similar to $A$ and it follows that $B^TB$ is similar to $A^TA$ too -- in fact, $B^TB=DA^TAD$ -- and the two matrices have identical spectra. Also, $A^TAx=\lambda x$ if and only if $B^TB(Dx)=\lambda Dx$. That is, $x$ is an eigenvector of $A^TA$ if and only if $y=Dx$ is an eigenvector of $B^TB$. So, your last assertion about the signs of $x$ and $y$ is true too.