When finding hybridisation of atoms, why are double bonds and triple bonds only considered once?
The hybridised orbitals are almost always involved in $\sigma$-bonding, and in housing the lone pair(s) of electrons. This helps us determine the geometry of the molecule.

Consider acetylene, $\ce{C_2H_2}$.

The carbon is $\mathrm{sp}$-hybridised. So, it has 2 $\sigma$-bonds - one to a hydrogen, one to another carbon. All this is in the third shell. So, a total of 4 orbitals (1 $\mathrm{s}$, 3 $\mathrm{p}$) undergo hybridisation and form 2 $\mathrm{sp}$-orbitals and 2 $\mathrm{p}$-orbitals. The $\pi$-bonding occurs due to these unhybridised $\mathrm{p}$-orbitals.

Hope this clears your doubt!

PS: The only exception to the first statement that I know is https://en.wikipedia.org/wiki/Aryne.