Suppose I know that $f\in\mathcal{C}^{k}(\mathbb{R})$ is such that there exist monic polynomials $P_{1}$, $P_{2}$ of order $k$, that bound $f$:
\begin{align*} \forall x\in\mathbb{R}, \ C_{1}P_{1}(x)\le f(x)\le C_{2}P_{2}(x)\end{align*}for some constants $C_{1}, C_{2}>0$. Can we then say for some perhaps different polynomials $\tilde{P}_{1}$, $\tilde{P}_{2}$, also of degree $k$, that
\begin{align*} \forall x \in\mathbb{R}, \ C_{1}|\tilde{P}_{1}(x)|\le |f(x)|\le C_{2}|\tilde{P}_{2}(x)|?\end{align*}
No. Let $f(x)=x+2\pi \sin x$ and $k=1$. Then the first condition holds with $P_1(x)=x-2\pi$, $C_1=1$, $P_2(x)=x+2\pi$, $C_2=1$. Assume there are $\tilde P_1$, and/or $\tilde P_2$ as requested.
We have $f(\pi)=\pi>0$, $f(\frac32\pi)=-\frac\pi2<0$, hence $f(x_0)=0$ for some $x_0\in(\pi,\frac32\pi)$. By symmetry, also $f(-x_0)=0$. Then $0\le C_1|\tilde P_1(\pm x_0)|\le |f(\pm x_0)=0$ implies that $\tilde P_1$ has two roots, contradicting $\deg \tilde P_1=k=1$. Hence no suitable $\tilde P_1$ exists.
Assume $\tilde P_2=x+b$ (you didn't specify explcitly that $\tilde P_2$ should be monic like $P_1, P_2$, but without that restriction the inclusion of $C_2$ in the inequality makes no sense). Then $-b$ msut be a zero of $f$. At that zero, $f'(-b)=1\pm 2\pi$, so $|f(-b+h)|\approx (2\pi-1)|h|$ for small $h$ whereas $C_2|\tilde P_2(-b+h)|=|h|$. For small $h$ this is a contradiction.