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 $K$.
What is true is that every extreme point of $M$ is an extreme point of the intersection of $K$ with one hyperplane, and this is a convex combination of two extreme points of $K$. Namely, suppose $p = \sum_{i=1}^r t_i p_i$, $t_i \in (0,1)$, $\sum_i t_i = 1$, is a convex combination of $r > 2$ extreme points of $K$. Say $T(p) = c$. If any $T(p_j) = c$, then $p$ is a convex combination of $p_j$ and $(p - t_j p_j)/(1-t_j)$ which are both in $M$, so not an extreme point. Otherwise some $T(p_i) > c$ and some $< c$. Relabelling, suppose $T(p_1) > c$ and $T(p_2) < c$. Then $$q = \frac{T(p_1) - c}{T(p_1)-T(p_2)} p_2 + \frac{c - T(p_2)}{T(p_1)- T(p_2)} p_1$$ is a nontrivial convex combination of $p_1$ and $p_2$ which is in $M$, and $p$ is a convex combination of this and some other member of $M$, thus not an extreme point.