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 $G$ is compact, its image under a character is discrete in the circle, hence finite, and the two concepts coincide.   
This applies in particular to profinite groups (since they are automatically compact).