I don't think this is quite right. Here is the right statement: let E_1, ..., E_g be the tails, with maps q_i: E_i --> B. Then

pi_* \omega_{C/B} = \bigoplus (q_i)_* \omega_{E_i/B}.

So, if your tails don't vary with B, this bundle is trivial.

Explanation: omega_{C/B} can be described explicitly: a section of omega_{C/B} is a one-form on each component of C, with simple poles at the nodes of C, so that at every node the residues of the form on the two components match.

Now, on a curve of genus 1, a one-form with only a single simple pole, must in fact have no poles. So the sections of omega_{C/B}, restricted to the E_i, are sections of omega_{E_i/B}. Moreover, the sections of omega_{C/B} restricted to the rational components are one-forms with no poles, and are hence 0. So to give a section of omega_{C/B} is simply to give a section of omega_{E_i/B} on each E_i. QED.