Solution:
Let $x_{1}, x_{2}$ be the zeros of $f(x)$, and let $y_{1}, y_{2}, y_{3}$ be the zeros of $g(x)$.
By Viete's relation,
$$\begin{aligned} x_{1} + x_{2} & = 2 r \\ x_{1} x_{2} & = r \end{aligned}$$
and
$$\begin{aligned} y_{1} + y_{2} + y_{3} & = r \\ y_{1} y_{2} + y_{2} y_{3} + y_{3} y_{1} & = \frac{s}{27} \\ y_{1} y_{2} y_{3} & = \frac{r^{6}}{27} \end{aligned}$$
Note that
$$\begin{gathered} \left(\frac{x_{1} + x_{2}}{2}\right)^{2} \geq x_{1} x_{2} \quad \Rightarrow \quad r^{2} \geq r \\ \frac{y_{1} + y_{2} + y_{3}}{3} \geq \sqrt[3]{y_{1} y_{2} y_{3}} \\ \frac{r}{3} \geq \sqrt[3]{\frac{r^{6}}{27}} \\ r \geq r^{2} \end{gathered}$$
Hence $r = r^{2}$, and consequently $x_{1} = x_{2}$ and $y_{1} = y_{2} = y_{3}$. Moreover, $r = 0, 1$.

- If $r = 0$, then $f(x) = x^{2}$ with $x_{1} = x_{2} = 0$. And since $y_{1} = y_{2} = y_{3}$ with $y_{1} + y_{2} + y_{3} = 0$, then ultimately $s = 0$.

- If $r = 1$, then $f(x) = x^{2} - 2 x + 1 = (x - 1)^{2}$ with $x_{1} = x_{2} = 1$. And since $y_{1} = y_{2} = y_{3}$ with $y_{1} + y_{2} + y_{3} = 1$ then $y_{1} = y_{2} = y_{3} = \frac{1}{3}$. Therefore $s = 9$.

Thus, the possible ordered pairs $(r, s)$ are $(0, 0)$ and $(1, 9)$.