Why RSA-KEM is more secure than Textbook-RSA?
Why is this not a problem?


Because for two different encryptions the random integers are drawn independently and uniformly at random over the whole range of the multipicative group $\mathbb Z_N^*$ (in practice this is usually approximated as $[1,n)$). The RSA assumption now literally states that it's difficult to recover the random value from its textbook RSA encryption if it is sampled in this way.


  if A and B again perform two key exchanges and it happens that $r_2 = 6943 r_1$ an attacker may be able to notice this and eventually even find out the random integers


Actually this won't happen because both $r_1,r_2$ are independently random, so is their group combination $r_1/r_2$ which has a chance of $2^{b-n}$ to be smaller than $2^b$ for $n$-bit RSA moduli, so for a really generous estimate suppose you are betting on $r_1/r_2$ to be guessable small, say smaller than $2^{60}$ and using weak 1024-bit RSA encryption, then you have a chance of $2^{60-1024}\approx 2^{-940}$ of the quotient being sufficiently small. This is a really negligible probability and likely won't ever be observed.