What is the role of the pyrrolidine/piperidine moieties in fagopyrins?
Radical generator

Looking at the molecules they are highly conjugated —must be dark red or similar!
And the disease is photosensitivity. Aromatics + UV? That is a standard UV radical reaction. It may be enough as a catalyst, but you'll get more will a cyclic amine, such as pyrrolidine or piperidine. If you want to go technical confer: Hofmann–Löffler reaction.

The radical will be passed on to other molecules, such as oxygen (generating peroxide), which will damage DNA and will raise an alarm biologically ultimately attracting neutrophils, which are expecting some attack (not poisoning), thus causing local inflammation, which manifests as skin irritation.

Demethylation

With the caveat that secondary metabolism is highly promiscuous and that some modifications are more accidental that anything else, let's look at the methylation and membrane permeability.

Promiscuity and secondary metabolism evolvability

This figure I found has the methylation patterns. A first glance I have assumed it was the usual C-methyltransferase action adding methyl groups, but instead it is a removal twice in the case of fagopyrin B and once in C and D. Specifically, fagopyrin is clearly two condensed anthraquinones molecules (emodin?) —which is a rather interesting non-standard change, I have not looked but so hope that someone has found the enzyme responsible! As you point out an anthraquinone is made by polyketide synthesis: the last acetyl-CoA results in a methyl group on the aromatic structure, which is occassionally oxidised to a carboxyl group (e.g. rhein) and in this case decarboxylated —I do not think a C-methyltransferase can remove groups even with pushed stoichiometries nor do I believe I have heard of one. But anyway. The anthraquinone condensing enzyme may not be able to tell apart the standard and the demethylated metabolites, which is otherwise meant for another pathway, i.e. it could just be promiscuous.

Membrane

Let's calculate the effect on solubility (python3 w/ RDKit).

Wikipedia has the SMILES string for the main fagopyrin and it's easy to spot the methyl groups.

from rdkit import Chem
from rdkit.Chem import AllChem
from rdkit.Chem.Draw import SimilarityMaps
# let's make the molecules
normal = Chem.MolFromSmiles('O=C1C2=C(C3=C(O)C(C4CCCCN4)=C2O)C(C5=C6C(C)=CC(O)=C15)=C7C(C3=C(O)C(C8NCCCC8)=C9O)=C9C(C%10=C7C6=C(C)C=C%10O)=O')
demeth = Chem.MolFromSmiles('O=C1C2=C(C3=C(O)C(C4CCCCN4)=C2O)C(C5=C6C=CC(O)=C15)=C7C(C3=C(O)C(C8NCCCC8)=C9O)=C9C(C%10=C7C6=CC=C%10O)=O')
# no need to orient them as the indices are the same. Whoohoo
img=Chem.Draw.MolsToGridImage([normal, demeth],
                              molsPerRow=2,
                              subImgSize=(200,200),
                              legends=['w/ methyls', 'w/o methyls'],
                              useSVG=False).save('/Users/matteoferla/Desktop/mols.png')




Now let's calculate logP (Partition coefficient).
    print('w/ methyls', Descriptors.MolLogP(normal))
    print('w/o methyls', Descriptors.MolLogP(demeth))


w/ methyls 6.71
w/o methyls 6.0


They are really soluble in octanol and insoluble in water. The unmethylated version is less hydrophobic.

Let's look at what is causing the difference by looking at the Crippen contributions to logP. Using the BluePurple colormap (suggestions welcome for best colormap for this descriptor), so that purple is hydrophobic while light blue is hydrophilic. Note that this is not partial charge (Gasteiger).

for mol, name in ((normal, 'meth'),(demeth, 'demeth')):
    AllChem.Compute2DCoords(mol)
    contribs = Chem.rdMolDescriptors._CalcCrippenContribs(mol)
    fig = SimilarityMaps.GetSimilarityMapFromWeights(mol, [x for x,y in contribs], colorMap='BuPu', contourLines=10)
    fig.savefig('/Users/matteoferla/Desktop/'+name+'_crippen.png', bbox_inches='tight')





The difference is a result of the methyl groups, being slightly more electron withdrawing than hydrogens making the carbon between it and the hydroxyl/ketone more hydrophobic. But overall does not change the polar sides, so is in effect unlikely to have an effect on its high membrane permeability.