C7 H10 O5

Which part is the functional group that could react with alcohol to form an ester ?

A carboxylic acid will "condense" with an alcohol to form an ester (water is formed in the reaction). So, the molecule must have a carboxylic acid group in it.

So am I right in thinking that O=C-O-H is the Carboxilic part that is the functional group that could react with alcohol to form an ester ?

That's correct. The "H" is "labile", meaning it can ionize. Thus it's an "acidic proton". The R-CO2-H "condenses" with OH-R' to form
R-CO2-R' + H2O. The C-H bond is replaced with a C-R bond.

I am now trying to draw shikimic acid... Showing all bonds

I have drawn a carbon chain one with a double bond.. I have done carbon to oxygen double bond and the same carbon has a oxygen -H

Next I have CH2 and although I know the rest of the molecule I'm struggleing to write it... C - O-H2 is that how I would write it?

Or would it be written C -H2- O ?

Im getting confused because the carbon has one single bond to the oxygen but the same carbon also has a single bond to a hydrogen...

Usually, for convenience, we write "-CH3" or "-CH2-" or the same sort of thing with the CH (I can't draw it here). Each carbon has four bonds, so we usually draw them left, right, up and down. You can show the H's separately, if you wish.

Most often, we use "line drawings" where the intersection of two lines is a carbon. The hydrogens are inferred. See this Wikipedia page for shikimic acid:

Shikimic acid - Wikipedia, the free encyclopedia

There's a line drawing at the right. The hexagon is inferring six carbon atoms with enough hydrogen atoms to fill four bonds to each carbon. Note the carbons next to the double bond. One of the carbons has no hydrogens connected to it because it already has four groups connected to it. The second carbon has but one hydrogen attached to it (inferred).