How can we prove that "We know that for any vector x, ${x}^{T}x=||x|{|}^{2}$. Thus , ..... "

The way I am thinking that while ${x}^{T}$ is the transpose of x, then we cross product with itself using ${x}^{T}$, which results in a symmetric matrix, R. So R is a square matrix. How can we jump from a square matrix to $||x|{|}^{2},$, where ||x|| supposed to mean the normal of x, then we square it with 2?

The way I am thinking that while ${x}^{T}$ is the transpose of x, then we cross product with itself using ${x}^{T}$, which results in a symmetric matrix, R. So R is a square matrix. How can we jump from a square matrix to $||x|{|}^{2},$, where ||x|| supposed to mean the normal of x, then we square it with 2?