Why RSA-KEM is more secure than Textbook-RSA? I am studying RSA-KEM but I can not see why it should be as secure as padding RSA messages (e.g with RSA-OAEP) when exchanging a key between two parties (say A and B).

It is easy to see that Textbook-RSA is problematic, as sending a key with $k_1^{e_B} \bmod n$ leaves the possibility for an attacker to analyze the message. Let's say that by coincidence A sends to B a second key $k_2^{e_B} = (42 k_1)^{e_B} = 42^{e_B} k_1^{e_B}$ (in the following $\bmod n$ is always omitted). An attacker could easily notice this as it could divide $k_2^{e_B}$ by $42^{e_B}$.

Now let us suppose that A and B are using RSA-KEM to exchange keys. An attacker cannot detect anymore if e.g. $k_2 = 42 k_1$ as a key encapsulation key is used to encrypt the shared key. However, the random integer to generate the key encapsulation key is still encrypted over Textbook-RSA. 

Why is this not a problem? An attacker is still able to perform e.g. statistical analysis on the randomly exchanged integer, e.g. 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, thus breaking the shared key as it depends on these integers.
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.