In the $s$-domain, the LPF-to-BPF transformation doubles the order of the filter. that is because the LPF has one transition from passband to stopband, but the BPF has two such transitions.
Remember that when we do frequency response, we substitute $s=j\omega$. That is, we evaluate the $s$-plane transfer function $H(s)$ as $H(j\omega)$. The magnitude of $H(j\omega)$ or $|H(j\omega)|$ is the gain of the filter when the input frequencey is $\omega$. Also keep in mind that the reciprocal of $j$ is its negative: $ \frac{1}{j} = - j $.
In the LPF, the middle of the passband is at DC, $\omega=0$. The transformation to BPF is such that maps, using a nice continuous function, the middle of the BPF passband at $\omega=\omega_0$ to $\omega=0$. A simple, mathematically continuous mapping as such is:
$$ \omega \leftarrow A ( \omega^2 - \omega_0^2 ) \quad A \ne 0 $$
So whenever the $\omega$ on the right (which is the BPF $\omega$) is equal to either $+\omega_0$ or $-\omega_0$, the $\omega$ on the left (which is the LPF $\omega$) is zero.
now suppose we set $A \triangleq \frac{Q}{\omega_0 \omega} $ which is still non-zero (and non-infinite for $\omega \approx \omega_0$). now you have
$$ \omega \leftarrow Q \left( \frac{\omega}{\omega_0} - \frac{\omega_0}{\omega} \right) $$
$$ \begin{align}
j\omega & \leftarrow Q \left( j\frac{\omega}{\omega_0} - j\frac{\omega_0}{\omega} \right) \\
& = Q \left( \frac{j\omega}{\omega_0} + \frac{\omega_0}{j\omega} \right) \\
\end{align} $$
or
$$ s \leftarrow Q \left( \frac{s}{\omega_0} + \frac{\omega_0}{s} \right) $$
so this means, regarding frequency response:
$$ \begin{align}
H_\text{BPF}(j\omega) & = H_\text{LPF}\left(j Q \left( \frac{\omega}{\omega_0} - \frac{\omega_0}{\omega} \right) \right) \\
& = H_\text{LPF}\left(Q \left( \frac{j\omega}{\omega_0} + \frac{\omega_0}{j\omega} \right) \right) \\
\end{align} $$
or, regarding transfer function:
$$ H_\text{BPF}(s) = H_\text{LPF}\left(Q \left( \frac{s}{\omega_0} + \frac{\omega_0}{s} \right) \right) $$
