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