Since you are looking to develop \$P\left(s\right)=\frac{u_2\left(s\right)}{u_1\left(s\right)}\$ in that circuit, it's not too complex. I assume you know that it's just a divider topology, so:

$$\begin{align*}
P\left(s\right)=\frac{u_2\left(s\right)}{u_1\left(s\right)}&=\frac{R_2\,\mid\mid\, L_1}{R_1+\left(R_2\,\mid\mid\, L_1\right)}\\\\
&=\frac{\frac{R_2\,\cdot\, s\,L_1}{R_2\,+\, s\,L_1}}{R_1+\frac{R_2\,\cdot\, s\,L_1}{R_2\,+\, s\,L_1}}\\\\
&=\frac{s\,L_1\,R_2}{R_1\,R_2+s\,L_1\left(R_1+R_2\right)}\\\\
&=\frac{s\frac{L_1}{R_1}}{1+s\frac{L_1}{R_1\,\mid\mid\, R_2}}
\end{align*}$$

If you set \$\omega_{_0}=\frac{R_1\,\mid\mid\, R_2}{L_1}\$ and set \$\sigma=0\$ (because you aren't interested in that part of \$s\$) so that \$s=j\,\omega\$, then this becomes:

$$\begin{align*}P\left(j\,\omega\right)&=\frac{j\,\omega\frac{L_1}{R_1}}{1+j\frac{\omega}{\omega_{_0}}}\\\\&=\frac{A\,j\frac{\omega}{\omega_{_0}}}{1+j\frac{\omega}{\omega_{_0}}}\text{, where }A=\frac{R_2}{R_1+R_2}\end{align*}$$

\$A\$ is usually interpreted as the voltage gain or attenuation (depending on its value, of course.) I think you can see where it comes from in the circuit, too.

The ***class*** of all \$1\$\$^\text{st}\$ order high pass filters can be completely analyzed by setting \$\omega_{_0}=1\$ and \$A=1\$, so that the simplifying high-pass filter is \$\frac{s}{1+s}\$. (All low-pass filters can be completely analyzed using the simplifying \$\frac{1}{1+s}\$.) See [digressions on passive filters](https://electronics.stackexchange.com/a/406893/38098) or, for putting a \$2\$\$^\text{nd}\$ order low-pass filter into standard form see [2nd order standard form](https://electronics.stackexchange.com/a/433573/38098).

This last link to the \$2\$\$^\text{nd}\$ order low-pass filter also provides a technique for generating a Bode plot by hand. It's for a low-pass, but the basic ideas remain. In a standard \$1\$\$^\text{st}\$ order high-pass filter, the corner angular frequency will be \$\omega_{_0}\$ (you can convert that to a frequency using the usual \$\omega_{_0}=2\pi\,f\$ and solving for \$f\$) and will be \$0\:\text{dB}\$ at higher frequencies (well, technically, no circuit goes to infinity as other parasitics always make a high-pass filter into a band-pass filter.) However, in your case, you know that \$A=\frac{R_2}{R_1+R_2}\$ (which almost certainly isn't 1), so you know this won't be \$0\:\text{dB}\$ but instead will be \$20\cdot\operatorname{log}_{10}\left(A\right)\$. Prior to the corner frequency for a \$1\$\$^\text{st}\$ order high-pass filter, the decline as you move down in frequency (usually graphed leftward away from the corner frequency) will be \$20\:\text{dB}\$ per decade (frequency.) So the sloped line is easy to draw.