Vtyjkyuk

Let $Ain M_3$ some arbitrary positive definite matrices and

B=$beginbmatrix
1& 1& 0& 2\
2& -1& 1& 2\
1 & 0& 0& 1
endbmatrix$.

Conclude connection between eigenvalue of matrices $B^TAB$, how much it has zero eigenvalues, how much it has positive and negative eigenvalues.

I find that $rankB=3$ then if I use some eigenvector $xin mathbb R^4$ such that $xnot=0$ then exist some eigenvalue $lambda$ that $B^TABx=lambda x$, if I multiply both side with $x^T$ I get this $x^TB^TABx=lambda x^Tx$
then if $Bx=y$ and if I write $x^TB^TABx=lambda x^Tx$ like this $y^TAy=lambda x^Tx$ then since $y^TAy>0$, $x^Tx>0$ so then $lambda>0$ but If I pick some eigenvector from $kerB$ then $lambda=0$ so it can be one zero and three positive eigenvalue, what you think?

From $A = Q Lambda Q^top$ with $Lambda$ a diagonal matrix we know that $A$ is self-adjoint. Thus, we know that all eigenvalues are real. Moreover, $B^top A B$ is self-adjoint:
$$
(B^top A B)^top = B^top (B^top A)^top = B^top A^top B = B^top A B
$$
Thus, the eigenvalues of $B^top A B$ are also real. Moreover, since $A$ is positive definite, $B^top A B$ is positive semidefinite and we know that the eigenvalues are all nonnegative, as you showed. This would be the formal argument.

Your remaining analysis is correct up to typos.