One-Time Pad Distribution of Keys Between Two Parties Consider the One-Time Pad (OTP). Suppose two parties, A and B, generate a completely random secret key in a real-life meeting, and they keep this secret. Now A wants to send some message to B using the OTP. The point is that in the OTP we can use any key only once. So we have to come up with a system to share a newly chosen key in a secret way.

Why cannot A just concatenate a new randomly generated secret key (needed for the next communication between the two parties using the OTP) to this message? 

For instance, A could concatenate this new key at the beginning of the message.
(Supposing they have agreed upon a fixed key-length beforehand, so that B is able to tell which is the new key and what was the "actual" message.) In this way A and B would agree on a new secret key which they can then use for their next communication. 

Probably this is insecure, but I do not immediately see why…
Let's say Alice and Bob have $c$ bits of pre-shared secret key material.  Alice generates $a$ bits of new key material, concatenates it to a message $M$ that is $b$ bits long, and uses $a+b$ bits of their pre-shared key material to encrypt "message||new-key".  She sends this secure message to Bob, who decrypts it with the shared key.  Now they both have $a$ bits of new shared key-material, but in order to securely send those $a$ bits and the $b$ bit message $M$ they had to burn through $a+b$ bits of the previous key material.  The total key material they now have to encrypt future messages is:

$c-$ (Amount spent to send "message||new-key") + $a$.  

The term (Amount spent to send "message||new-key") = $a+b$, so the amount they have left is:

$c-(a+b)+a$,

$= c-a-b+a$,

$= c-b$.

So they are no better off now than if Alice had simply send the $b$ bit message $M$ alone without an appended new key.