According to Wikipedia 4-methyl-1-pentanol is $\ce{C_6H_{14}O}$.
But here is what I got, which is slightly different.
Methyl = Think of methyl group, which is $\ce{CH_3}$.
Pentanol = Penta means we have 5 carbons, -ane ending means we have $2n+2$ hydrogens, i.e. 12 hydrogens and -ol ending means we have a hydroxyl group $\ce{-OH}$.
So the molecular formula that I arrived at is: $\ce{CH_3C_5H_{12}O}$ = $\ce{C_6H_{15}O}$.
How did I end up with an extra hydrogen compared to Wikipedia answer?
Actually, the formula you arrived at is $\ce{CH3C5H12OH}$ = $\ce{C6H16O}$. In any case, in your "2n+2" step, you need to reduce that number of hydrogens by 2. One is eliminated where the methyl group is attached, and a second is eliminated where the $\ce{-OH}$ group is attached. So removing 2 hydrogens from $\ce{C6H16O}$, gives you the desired $\ce{C6H14O}$.
You’re applying the $2n +2$ method wrongly. You need to stupidly count all the carbons in a fully saturated molecule and take that number as $n$.
In your case this is:
That gives you six carbons in total. $n = 6 \Rightarrow 2n + 2 = 14$
Alcoholic oxygens do not introduce unsaturation, so you can safely ignore the oxygen here.
