Let $\beta=2^{1/4}$, and let $\alpha=\beta^3+\beta^2+1$. Then ${\bf Q}(\alpha)={\bf Q}(\beta)$ has degree 4 over the rationals. Then $\alpha^2=2\beta^3+4\sqrt2+4\beta+3$, and there is no quadratic polynomial $f$ with rational coefficients such that $f(\alpha)$ has degree 2 over the rationals.

Let $\beta=2^{1/4}$, and let $\alpha=\beta^3+\beta^2+1$. Then ${\bf Q}(\alpha)={\bf Q}(\beta)$ has degree 4 over the rationals. Then $\alpha^2=2\beta^3+4\sqrt2+4\beta+3$, and there is no quadratic polynomial $f$ with rational coefficients such that $f(\alpha)$ has degree 2 over the rationals.

For another example, let $\alpha$ be a primitive 5th root of unity. The quadratic subfield of ${\bf Q}(\alpha)$ is real, so if $f(\alpha)$ is in it then it equals its complex conjugate. For $f$ quadratic, this leads to a degree 3 equation for $\alpha$, contradiction.