Rather than the frequency domain, let's look at this in the time domain and particularly, the characteristic equation associated with a linear homogeneous 2nd order differential equation for some system:
\$r^2 + 2 \zeta \omega_n r + \omega^2_n = 0\$.
If the roots of the characteristic equation are real (which is the case if \$\zeta \ge 1\$), the general solution is the sum of real exponentials:
\$Ae^{\sigma_1 t} + Be^{\sigma_2t} \$
where
\$\sigma_1 = -\zeta \omega_n + \sqrt{(\zeta ^2 - 1)\omega^2_n} \$
\$\sigma_2 = -\zeta \omega_n - \sqrt{(\zeta ^2 - 1)\omega^2_n} \$
Since these are real exponentials, there is no oscillation in these solutions.
If the roots are complex conjugates (which is the case if \$\zeta < 1\$), the general solution is the sum of complex exponentials:
\$e^{\sigma t}(Ae^{j\omega t} + Be^{-j\omega t})\$
where
\$\sigma = -\zeta \omega_n\$
\$\omega = \sqrt{(1 - \zeta ^2)\omega^2_n}\$
This solution is a sinusoid with angular frequency \$\omega\$ multiplied by a real exponential. We say the system has a "natural frequency" of \$\omega\$ for a reason that I think is obvious.
Finally, setting \$\zeta = 0\$ (an undamped system) , this solution becomes:
\$Ae^{j\omega_n t} + Be^{-j\omega_n t}\$
which is just a sinusoid of angular frequency \$\omega_n\$.
In summary, a system may or may not have an associated natural frequency. Only systems with \$\zeta < 1\$ have a natural frequency \$\omega\$ and only in the case that \$\zeta = 0\$ will the natural frequency \$\omega = \omega_n\$, the undamped natural frequency.
You want to use the following property of Laplace transform:
$$\mathscr{L}\left({\frac{dx(t)}{dt}}\right)(s) = s\mathscr{L}\left(x(t)\right)-x(0)$$
This allows you to easily move between differential equations and polynomial equations.
Time domain to frequency domain: Take your first equation for example
$$\frac{d^2 x(t)}{dt^2} + 2 \zeta \omega_n \frac{dx(t)}{dt} + \omega_n^2 x(t) = f(t).$$
If we denote the Laplace transform of \$x(t)\$ by \$X(s)\$ and \$f(t)\$ by \$F(s)\$ and apply Laplace transform to this equation then this property implies
$$s^2 X(s) + 2 \zeta \omega_n sX(s) + \omega_n^2 X(s) = F(s)$$
where for simplicity I assume that \$x(0) = f(0) = 0\$.
The transfer function is defined as the ratio:
$$\frac{X(s)}{F(s)} = \frac{1}{s^2+2 \zeta \omega_n s + \omega_n^2}$$
Frequency domain to time domain: Lets try the example \$G(s)=\frac{1}{s^3+1}\$ then we have by definition that
$$G(s)F(s) = X(s)$$
which implies
$$F(s) = s^3X(s)+ X(s).$$
Taking inverse Laplace transform we find that
$$f(t) = \frac{d^3 x(t)}{dt^3} + x(t).$$
Hopefully this allows you to see the pattern in general.
Best Answer
For a second-order system featuring a double zero, the transfer function could look that way:
$$ H(s)= H_0\frac{1 + a_1s + a_2s^2 }{ 1 + b_1s + b_2s^2} = H_0 \frac {N(s)}{D(s)}$$
you can rearrange both \$ N(s) \$ and \$ D(s) \$ under the form
\$ N(s) = 1 + (s/\omega_{0n}*Q_n) + (s/\omega_{0n})^2 \$ and \$ D(s) = 1 + (s/\omega_{0}*Q) + (s/\omega_{0})^2 \$
with \$ \omega_{0n} = 1/\sqrt{a_2} \$ and \$ Q_n = \sqrt{a_2}/a_1 \$ with \$ \omega_{0} = 1/\sqrt{b_2} \$ and \$ Q = \sqrt{b_2}/b_1 \$
In your first expression, factor \$ K_3 \$ on top and \$ K_5 \$ in \$ D(s) \$, you have
$$ H(s) = \frac{K_3}{K_5}\frac{(1 + \frac{sK_2}{K_3} + \frac{s^2K_1}{K_3}) }{ 1 + \frac{sK_4}{K_5} + \frac{s^2}{K_5}} $$ then apply what I described above.
Same for the second expression:
$$ H(s)= \frac{K_2}{K_4} \frac{(1 + sK_1/K_2) }{1 + \frac{sK_3}{K_4} + \frac{s^2}{K_4}} $$ then apply what I described above.
If you want to learn more about these techniques, have a look at the APEC presentation available here:
http://cbasso.pagesperso-orange.fr/Downloads/PPTs/Chris%20Basso%20APEC%20seminar%202016.pdf