Standard parameters of a second order transfer function with some zeros
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