I find the statement confusing, because the conclusion is not as strong as it should be. I think the conclusion should be as follows:

> If $\gamma$ is a path in $\Sigma$ whose endpoints are fixed by $T$,
> and if $T \circ \gamma$ is homotopic to $\gamma$ (rel endpoints), then there is a homotopy of $\gamma$ (rel endpoints) making $\gamma$ disjoint from the union $\cup C_i$.

The proof is as follows: Homotope $\gamma$ (rel endpoints) to have minimal intersection number with $\cup C_i$.  If the result is disjoint from $\cup C_i$.  If not then $\gamma$ crosses one of the curves, say $C$.  Let $(B_j)$ be the subcollection of $\cup C_i$ which are parallel to $C$.  Let $X \subset \Sigma$ be an annulus that contains all of the $B_j$ as essential curves in its interior.  We may also assume that $\gamma$ has minimal intersection number with $\partial X$.  Let $\delta$ be a component of $\gamma \cap X$ which meets some of the $B_j$. We may (and do) assume that the endpoints of $\delta$ are fixed by all of the $T_i$. 

Let $\Sigma^C$ be the cover of $\Sigma$ which is homeomorphic to an annulus and where $C$ does not unwrap.  We use the initial point of $\delta$ as our base-point; let $\gamma^C$ be the resulting lift of $\gamma$ to $\Sigma^C$.  Note that $\delta$ thus lifts to $\delta^C \subset \gamma^C$.  The arc $\delta^C$ that crosses the lifts of (the correct) $B_j$ due to our choice of base-point. 

We now lift $T(\gamma)$ using the *same* base-point.  It lifts (up to homotopies in $\Sigma^C$ supported in bigons disjoint from $C$) to a copy of $T_C^n(\gamma^C)$.  Here $n$ is the number of parallel copies of $C$ crossed by $\delta$ (and the sign depends on the sign of the twists along the $B_j$).  Since $T_C^n(\gamma^C)$ has different winding number (than $\gamma^C$) with the curves $B_j$, it follows that they are not homotopic (rel endpoints) in $\Sigma^C$.  Thus (by homotopy lifting) $T(\gamma)$ and $\gamma$ are not homotopic (rel endpoints) in $\Sigma$.