It's called partial fraction decomposition.
$\rm \frac{1}{(as+b)(cs+d)}=\frac{\square}{as+b}+\frac{\triangle}{cs+d}$
Setting up constraints on $\rm \square,\triangle$ to make the above true for $\rm s$ arbitrary,
$\rm (c\square+a\triangle)s+(d\square+b\triangle)=0s+1 \implies\quad \begin{cases} \rm c\square+a\triangle=0 \\ \rm d\square+b\triangle=1\end{cases}$
which is a linear system with solution
$\rm \begin{pmatrix}\square \\ \triangle\end{pmatrix}=\begin{pmatrix}c & \rm a \\ \rm d & \rm b\end{pmatrix}^{-1}\begin{pmatrix}0 \\ 1\end{pmatrix}=\frac{1}{ad-bc}\begin{pmatrix}a \\ \rm -c\end{pmatrix}.$
There is a much simpler case, though:
$\rm \frac{1}{s(s+h)}=\frac{1}{h}\frac{h}{s(s+h)}=\frac{1}{h}\frac{(s+h)-s}{s(s+h)}=\frac{1}{h}\left(\frac{1}{s}-\frac{1}{s+h}\right).$
Note that the most general form can be brought into this form:
$\rm \frac{m}{(as+b)(cs+d)}=\frac{m}{ac(s+b/a)(s+d/c)}=\frac{m}{ac}\frac{1}{r(r+d/c-b/a)},$
where $\rm r=s+b/a$. Thus we have
$\rm \frac{3}{s(0.1s+1)}=\frac{30}{s(s+10)}=\frac{30}{10}\left(\frac{1}{s}-\frac{1}{s+10}\right)=\frac{3}{s}-\frac{3}{s+10}.$
(There is indeed a factor of $3$ missing in the notes, apparently.)