Is this concatenation of two FA done right? $r_3=r_1r_2=(a^*b)^*(a+ba)^*bb(a+b)^*$ comes out to be $r_3=r_2=(a+ba)^*bb(a+b)^*$ when i generate the resultant FA and its regex after concatenation i.e. it doesn't include $r_1$
Details:
Consider these two FA
$FA_1$

$FA_2$

if $r_1,r_2$ and $r_3$ are regex for $FA_1,FA_2$ and $FA_3$ respectively where
$r_1=(a^*b)^*$
$r_2=(a+ba)^*bb(a+b)^*$
then, mathematically, $r_3=r_1r_2=(a^*b)^*(a+ba)^*bb(a+b)^*$
But $r_3$ turns out different when I actually concatenate $FA_1$ and $FA_2$ and use the new transition table (given below) for $FA_3$ to generate its regex ($r_3$) and the $TG_3$ (given below) and then simplify, $r_3$ becomes $(a+ba)^*bb(a+b)^*$ which is the same as $r_2$ (i.e. $r_1$ not prefixed). and since $r_3=r_1r_2=(a^*b)^*(a+ba)^*bb(a+b)^*$ can't be simplified into $r_2=(a+ba)^*bb(a+b)^*$
I am convinced that the problem is in the concatenation/transition table,
especially with the circular references to $x_1$ and $y_1$, which i don't completely get. That also would mean $TG_3$ iswrong.  

The Transition Table:
   

$TG_3$


Now, it'd be very tedious to explain how i created the transition table, it's easier to figure out by looking at it. but key point i can't quite grasp the "$x_1$ and $y_1$ being connected" part
Sidenotes:
 - This is the only way to concatenating two FA that i know of. I am on a distant learning platform and it's quite low quality material, so links to proper methods are appreciated.
 - If you use another method, please keep in mind that the transition table, regex and the TG and all required
 - The TG looks like an FA to me (barely know the difference) but the site i used to generate it says it's a TG. what's the difference?
You can think of the concatenation of the languages defined by the two automata as the language of the automaton that can be obtained by connecting all final states of the first (in the example, just $x_1$) with the starting state of the second ($y_1$), by an empty transition, and then making them non-final. 

A transition table can then be used to convert the NFA just generated into a DFA, which is what you did.

In this case, $r_1r_2$ is indeed equivalent to $r_2$, and this is completely normal. A transition graph represents a transition function, which is part of an automaton. If you minimize $TG_3$, what you get is exactly $FA_2$, which confirms the equivalence.