Suppose that given is: $A_{x}=\begin{bmatrix} x & -1 \\ -1 & x\\ \end{bmatrix}$
What is the signature of $A_1$ and $A_2$? Is it simple as I think: (1,0,1) and (0,1,1), because the first one has eigenvalues 0 and 2 and the second 0 and -2?
And another question: If $B:V\times V\to R$ symmetric and bilinear and for all $x,y$ in V: $B(x,x)\geq0$ and $B(y,y)\geq0$, does $B(x+y,x+y)\geq0$ applies also then?
Yes it diagonalizes to $\begin{bmatrix}x-1&0\\0&x+1\end{bmatrix}$, and from that you can read the eigenvalues {0,2} and {1,3}, indicating signatures of 0+ and ++. It looks like you may have miscomputed the eigenvalues.
I think in the second part you are trying to ask "if B(x,x) and B(y,y) are both nonnegative, then must B(x+y,x+y) also be nonnegative?" This isn't the case. Hint: find two null vectors x,y with B(x,y) nonzero. If B(x,y) isn't already negative, replace x with -x. This furnishes a counterexample.