quantum field theory - Time-ordered operator in Srednicki

On page 51 Srednicki states, "Note that the operators are in time order...we can insert $T$ without changing anything". This I agree with. But then on the next paragraph he states "The time order operator $T$, moves all...where they annihilate...". How correct is this paragraph? Is equation (5.14) always valid? Is equation (5.14) the reason why the time-ordering symbol $T$ appears all over the place in QFT?

The equations I'm referring to are (braket package available?):

$$\langle f|i\rangle = \langle0|~a_{1'}(+\infty)a_{2'}(+\infty)a_{1}^\dagger(-\infty)a_{2}^\dagger(-\infty)~|0\rangle \tag{5.13}$$

and since time goes from right to left inside the braket we put the $T$-symbol

$$\langle f|i\rangle = \langle 0|T ~a_{1'}(+\infty)a_{2'}(+\infty)a_{1}^\dagger(-\infty)a_{2}^\dagger(-\infty)~|0\rangle \tag{5.14}$$

without changing anything.

After this, the $T$-symbol appears everywhere!