Ok, lets see if I can explain this correctly... for hybridization, just count all the bonding and nonbonding electron pairs of the molecule. If you don't know how many they have, you have to draw the Lewis structure!
Double and triple bonds count as one. So, if I have a BF3 molecule that is trigonal planar, it has three bonding pairs. It has no nonbonding electron pairs. 3 + 0 = 3 This means the hybridization will be sp2.
Same would go for a bent molecule with 2 bonds and 1 non bonding electron pair. Together that is three. So 3 is sp2.
For tetrahedral, there are four bonds. Four is always sp3
Same would go for trigonal pyramidal, because there are 3 bonds, then 1 nonbonding electron pair, which together is 4.
You might see a pattern that a certain shape will have a certain hybridization. But this is NOT true! For example a bent molecule can have 2 bonds and 2 nonbonding electron pairs. Together that is 4. And what is 4? sp3!
Alternatively you could memorize the hybridization by things like "AX5E1 is always sp3d2" but I feel that's unnecessary
How to get the numbers?
Add the bonds and nonbonding electron pairs together.
Double and triple bonds count as ONE.
One last example:
Try drawing the structure for I3^-1
You will see that it is linear. It has 2 bonds. It has 3 nonbonding electron pairs. 2 + 3 = 5. What is 5? sp3d2.
I'm sure there are a bunch of another ways but I find this the simplest and I hope I helped!