In quantum mechanics how the expression of average value of an observable is derived?

Maybe I could help you with your confusion. If we consider two hermitian operators (just ordinary QM) $A,B$ and regard $$<v|AB|v>$$ you could run into several scenarios.

1. $|v>$ is eigenstate for operator $A$ (with eigenvalue $a$) and operator $B$ (with eigenvalue $b$). Then your argumentation is correct. $<v|AB|v>=ab<v|v>=<v|A|v><v|B|v>$

2. If $|v>$ is not eigenstate to one of those operators (or both) then $<v|AB|v>=<v|A|v><v|B|v>$ does no longer hold in general. We could consider one counter expamle: $$A = begin{bmatrix}0 & 11 & 0end{bmatrix}$$ $$B = A$$ $$AB = begin{bmatrix}1 & 0 & 1end{bmatrix}$$ $$|v> = begin{bmatrix}1end{bmatrix}$$ Then : $$<v|AB|v>=begin{bmatrix}1&0end{bmatrix}begin{bmatrix}1 & 0 & 1end{bmatrix}begin{bmatrix}1end{bmatrix}=1$$ But : $$<v|A|v>=begin{bmatrix}1&0end{bmatrix}begin{bmatrix}0 & 11 & 0end{bmatrix}begin{bmatrix}1end{bmatrix}=0$$

In fact, this is very important to realise that the operators are not allways diagonal.

It is generally NOT true that $B|xrangle=b|xrangle$. This is only true if $|xrangle$ is an eigenvector of $B$. However, there is another flaw in your reasoning. Even if $|xrangle$ is an eigenvector of both $A$ and $B$ (as you assumed), it is NOT true that $langle x| xrangle=1$ (I assume $|xrangle$ represents an eigenvector of the position operator $X$). Position eigenvectors are not normalizable in the usual sense. They are normalized to the dirac delta function such that $langle y|xrangle=delta(y-x)$, which means $langle x|x rangle=delta(0)$ in infinite.