The Paper [1] says in definition 2.1
suppose E has a K-rational point of order r
Lets assume $K=\mathbb F_p$ for any prime $p$. For this case I could consider the r-Torsion group on E, say $T_{r,E}:=\{P\in E\ \mid\ rP=\mathcal O\}$, then I guess, the set of K-rational points with order r on E is equal to $T_{r,E}$. Am I right or completely wrong?
I guess the right point to start with, is [2] (and a bid up the general definition). Consider $y^2=f(x)$ defines an elliptic curve. Then the definitions says: The $K$-rational Points of E is defined as the set $R:=\{(x,y)\in K^2\ \mid\ f(x)=0\}$. That means, any point on the curve, reduced to $K$ is a $K$-rational point. If I consider the case of "degree r", it turns out, that the set is equal to $\{P\in R\ \mid rP=\mathcal O\}$. Which is the origin of the r-Torsion subgroup of E. (This is what I guess to be right with.)
References
[1] http://theory.stanford.edu/~dfreeman/papers/taxonomy.pdf