It is easy to miss the point  that in the second definition $\mathbb Q/\mathbb Z$ is required  to be *discrete* in  Hiro's question.   
Hence even if a  continuous morphism $f:G\to T$ has image in $\mathbb Q/\mathbb Z$ it cannot automatically be considered as a continuous map $f_0:G\to \mathbb Q/\mathbb Z$  and so might not be a character in the second sense.  

An example for this failure is to take $G=T_{tors} =\mathbb Q/\mathbb Z\subset T$ , the torsion subgroup of $T$ with its induced topology from the circle and for the character $f $  (in the first sense) the inclusion $f:G\hookrightarrow T$.   
Even though $G$ is torsion, the corestricted morphism $f_0:G\to \mathbb Q/\mathbb Z$ is not a character in the second sense, since it is not continuous.   

However if a character on  a compact group $G$  *happens*  to have values  in $\mathbb Q/\mathbb Z$,so that both definitions can be compared,  then its image  in the circle is finite and the two concepts coincide.       
Xandi explains in his answer  that this is always the case for profinite groups.