Dissolving NaOH into Methanol will give 18g of water for each 40g of NaOH - assuming 100% pure and 100% conversion.
With KOH the figures are 18g of water from 56g of KOH - same assumptions.
With most reactions the process is never 100% complete but sits somewhere in the middle, in the case of say aqueus calcium hydroxide and sulphuric acid this process is very much towards the production of calcium sulphate (insoluble precipitate), where as with aqueous sodium hydroxide and sulphuirc acid it is very much in the middle as everything separates into the ions which all remain in solution.
With KOH (or NaOH) dissolving into methanol the reaction can be viewed as:
KOH + MeOH <--> KOMe +H2O
How much will be KOH vs how much is KOMe will depend on a number of factors, such as concentration, temperature, any contaminents, possibly even the material the container is made of.
If one of the compounds is removed, for eaxmple KOMe by transesterification of veg oil or KOH by making soap with the veg oil, then that would disturb the equilbrium so moving it more to one direction, as a result all other things being equal the equilibrium will be reestablished by converting more KOH to KOMe or KOMe to KOH.
This also means that if any attempt to measure the water level is performed during the reaction (transesterification as far as we are concerned), it may appear to increase water level as more KOH is converted to KOMe because the KOMe is used up. Problems could lie in the fact that glycerol absorbs water which could adversely impact any test results as this may or may not be reflected in the results.