I believe that I have the answer. I would like to hear what people think.
The objective here is to prove $\{x,Tx,\ldots,T^{M-1}x\}$ lie in the same orbit segment $\{ S^j(y)\}_{j=0}^\infty$, for all but $\epsilon$ of the space.
Given a ladder $\mathcal{L} = \mathcal{L}(\{C(i)\},N,S)$. For any $x \in X$ there exists $y_0 \in C(0)$ such that $x = S^{j_0}(y_0) = T^{n_{y_0}}(y_0)$ with $j_0 < N$.
Similarly, for $T^kx \in X$, $k \in \{1,\ldots,M\}$ there exists some $y_k \in C(0)$ such that $T^kx = S^{j_k}(y_k) = T^{n_{y_k}}(y_k)$ with $j_k < N$. So
$$ T^{n_{y_k}}(y_k) = T^k(x) = T^{n_{y_0} + k}(y_0) $$
$$y_k = T^{n_{y_0} + k - n_{y_k}}(y_0)$$
That is to say, each of the $y_k \in C(0)$ can be placed in the $T$-orbit of $y_0$. Since $y_k \in C(0)$, we can further say that each of the $y_k$ can be placed in the same $T\vert_{C(0)}$ orbit of $y_0$.
Let $M_1$ be so large that for all but $\epsilon_1/M$ of the space the return time to $C(0)$ is less than $M_1$. Then for all but $\epsilon_1$ of the space, $n_{y_k} < M_1$ and $n_{y_0} + k - n_{y_k} < M_1 + k < M + M_1$
By Rohlin's lemma we can create $\mathcal{L}_0$ which (for most of the space) contains an orbit segment of arbitrary length. Choose this length to be $M + M_1$. This new ladder contains $\{ y_{k}\}_{k=0}^M$ as an orbit segment for all but $\epsilon_1$ of the space $C(0)$.
Refining $\mathcal{L}$ by $\mathcal{L}_0$ gives a new ladder $\mathcal{L}_1$. By construction, the orbit of $\{S_1^j(y_0)\}_{j=0}^\infty$ contain $y_k$ because $S_1^N = T \vert_{C(0)}$, and for any $k < M$
$$ T^k(x) = S^{j_k}(y_k) \in \{S^j(y_k)\}_{j=1}^N \subset \{S_1^j(y_0)\}_{j=0}^\infty $$