The last paragraph is wrong. Consider the case $m=1$. $\tilde{M}$ is the intersection of $K$ with one hyperplane. In general this will not contain any extreme point of $K$, so its extreme points will not be convex combinations of $m=1$ extreme points of $K$.
Or did you mean $n$ instead of $m$? Krein-Milman theorem says any compact convex subset of a locally convex tvs is the closed convex hull of its extreme points; in this case $K$ has at most $n$ extreme points so the convex hull of these is closed. Thus every point of $K$, and in particular every extreme point of $M$, is a convex combination of the extreme points of $K$.