Closure of regular Language - Transition Function : Sipser Proof
The idea is that whenever $A_1$ reaches an accepting state, it "guesses" that this is a good place to split the word, and run the second part in $A_2$. But maybe this guess is wrong? There are many ways to split a word, and you only need one such splitting to work.

As an example, consider the languages $L_1=(a+b)^*a$ and $L_2=b(a+b)^*$. Now consider what would happen in your suggestion with the word $aab$.
After reading the first $a$, $A_1$ will be in an accepting state, so it will move to the initial state of $A_2$, and upon reading the second $a$, will move to a rejecting sink.

However, as Sipser suggests, $A_1$ may continue using $\delta_1$ for the second $a$ as well, and only then move to $A_2$. 

Same goes for the construction for Kleene star.