Is this concatenation of two FA done right?
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.