Let’s try to put everything in an order.
1). Organic reactions are not only nucleophilic substitution ones but also condensations, percyclic, redox reactions, etc. Thus, principles being valuable for nucleophilic substitution, might not always be valuable for other kinds of organic reactions. So, forget the nucleophilic substitution for the moment. The said reaction is condensation catalyzed by nitrile anion and accompanied by basic co-catalysis.
2). “Weaker the base, better it is the leaving group” is not true. The truth is, the stronger the nucleophile, the easier leaving of the nucleophyge group, is. The confusion starts by the aphorism that “the stronger the base, the stronger the nucleophile is” and its many exceptions, e.g. tertiary amines are weak to medium bases (of course, not all and thus, another exception of the exception) but they are strong nucleophiles and therefore, they easily form quaternary ammonium salts (except fluorides).
3). There is no formation of ROK in alcoholic KOH solution and the base is still KOH. But the lower the ionization of the dissolving medium, the higher the strength of acids and bases, is. Therefore, alcoholic KOH is a stronger base than aqueous KOH.
4). In the given reaction mechanism there will be a proton exchange of the intermediary carbanions and alkoxydes, if not strong basic conditions are and thus, the reaction would stop. Alcoholic KOH comes in equilibrium (indeed, localized to the left) with the corresponding intermediary carbanions and alkoxydes.
5). As you mentioned before: Simply cyanohydrin is formed by nucleophilic addition (to carbonyl, in aqueous medium and thus, the reaction would stop). Besides, formation of cyanohydrin or of cyanohydrin ether is very difficult in alcoholic medium.
6). For your one education, the mechanism of benzoin condensation is still not very well clarified, yet. Apart the given mechanism, two other mechanism models have also been proposed. The one assumes a more active role of the base rather than a co-catalyst and the other one assumes the formation of a dianion in the first step and notably, the formation of both the carbanion and the alkoxyde after the nitrile attack to benzaldehyde. But for the moment, forget them until graduation.
7). Is it more clear, now?