This cannot work. If a polynomial from $\mathbb R[\mathbf{x}]$ vanishes on $\mathbb R^n$, then it is $0$ itself. So if $P(\mathbf{v})^tP(\mathbf{v})=I_2$ for each $\mathbf{v}\in\mathbb R^n$, then $P(\mathbf{x})^tP(\mathbf{x})=I_2$. Similarly, we get $P(\mathbf{x})^tM(\mathbf{x})P(\mathbf{x})=D(\mathbf{x})$, so $M(\mathbf{x})$ has eigenvalues in $\mathbb R[\mathbf{x}]$. However, a matrix like $\begin{pmatrix}x_1^2&x_1\\x_1&0\end{pmatrix}$ doesn't have eigenvalues in $\mathbb R[\mathbf{x}]$ (for instance because the discriminant of its characteristic polynomial is $x_1^2(x_1^2 + 4)$).

This cannot work. If a polynomial from $\mathbb R[\mathbf{x}]$ vanishes on $\mathbb R^n$, then it is $0$ itself. So if $P(\mathbf{v})^tP(\mathbf{v})=I_2$ for each $\mathbf{v}\in\mathbb R^n$, then $P(\mathbf{x})^tP(\mathbf{x})=I_2$. Similarly, we get $P(\mathbf{x})^tM(\mathbf{x})P(\mathbf{x})=D(\mathbf{x})$, so $M(\mathbf{x})$ has eigenvalues in $\mathbb R[\mathbf{x}]$. However, a matrix like $\begin{pmatrix}1&x_1\\x_1&0\end{pmatrix}$ doesn't have eigenvalues in $\mathbb R[\mathbf{x}]$ (for instance because the discriminant of its characteristic polynomial is $1+4x_1^2$).

This cannot work. If a polynomial from $\mathbb R[\mathbf{x}]$ vanishes on $\mathbb R^n$, then it is $0$ itself. So if $P(\mathbf{v})^tP(\mathbf{v})=I_2$ for each $\mathbf{v}\in\mathbb R^n$, then $P(\mathbf{x})^tP(\mathbf{x})=I_2$. Similarly, we get $P(\mathbf{x})^tM(\mathbf{x})P(\mathbf{x})=D(\mathbf{x})$, so $M(\mathbf{x})$ has eigenvalues in $\mathbb R[\mathbf{x}]$. However, a matrix like $\begin{pmatrix}x_1&x_1\\x_1&0\end{pmatrix}$ doesn't have eigenvalues in $\mathbb R[\mathbf{x}]$.

This cannot work. If a polynomial from $\mathbb R[\mathbf{x}]$ vanishes on $\mathbb R^n$, then it is $0$ itself. So if $P(\mathbf{v})^tP(\mathbf{v})=I_2$ for each $\mathbf{v}\in\mathbb R^n$, then $P(\mathbf{x})^tP(\mathbf{x})=I_2$. Similarly, we get $P(\mathbf{x})^tM(\mathbf{x})P(\mathbf{x})=D(\mathbf{x})$, so $M(\mathbf{x})$ has eigenvalues in $\mathbb R[\mathbf{x}]$. However, a matrix like $\begin{pmatrix}x_1^2&x_1\\x_1&0\end{pmatrix}$ doesn't have eigenvalues in $\mathbb R[\mathbf{x}]$ (for instance because the discriminant of its characteristic polynomial is $x_1^2(x_1^2 + 4)$).