I ended up asking Jon Chaika this question, and he gave a rough sketch, in which I filled in the details, so if there are any mistakes, they are probably mine.
Suppose that there were $n+1$ measures $\{\mu_1, \ldots, \mu_{n+1}\}$ that were generic with respect to $T$ and $T^{-1}$, and let $\{x_1, \ldots, x_{n+1}\}$ be their generic points. Pick a subset $U$ that gets assigned different measure by each $\mu_i$ (it's possible that no single set will do, in which case we pick finite collection of subsets, but the idea is the same). Let $2 \varepsilon$ be the smallest difference in the size of the sets: we have that $2 \varepsilon > 0$.
Since each $x_i$ is $\mu_i$-generic with respect to both $T$ and $T^{-1}$, there exists a large enough time $t_i \in \mathbb{N}$ such that the average time the orbit of $x$ under both $T$ and $T^{-1}$ spend in $U$ for time $t \geq t_i$ is within $\varepsilon$ distance of $\mu_i(U)$.
\begin{align*}
\left|\frac{\#(\{T^{j}x_i\}_{j=0}^{t} \cap U)}{t} - \mu_i(U) \right| &\leq \varepsilon \\
\left|\frac{\#(\{T^{-j}x_i\}_{j=0}^{t} \cap U)}{t} - \mu_i(U) \right| &\leq \varepsilon
\end{align*}
Now pick a very small sub-interval of our original IET $T$, and consider the first return map to this smaller sub-interval. By making the subinterval sufficiently small, we can ensure that the first return time for any point in the subinterval is greater than $2 \max(t_i)$. Furthermore, since our original IET had $n$ intervals, the induced IET will also have $n$ intervals.
Label the $n$ intervals of the induced IET as $\{J_1, \ldots, J_n\}$ and let the return times of each of these intervals be $\{r_1, \ldots, r_n\}$. We make the following observations about the orbits of $J_i$:
$$T^aJ_c \cap T^b J_d = \emptyset$$
where $0 \leq a \leq r_c$, $0 \leq b \leq r_d$ and $a \neq b$ and $c \neq d$. This means that in the time period we're interested in, the orbits are disjoint. But in the very same time period, the orbits fill out $[0,1]$.
$$[0, 1] = \bigcup_{i=1}^{n} \bigcup_{a=0}^{r_i - 1} T^a J_i$$
By the pigeonhole principle, at least $2$ of the $n+1$ generic points must be in the same $J_i$ orbit. Without loss of generality, let's assume $x_1$ and $x_2$ lie in the orbit of $J_1$. More specifically, we have the following:
\begin{align*}
x_1 &\in T^{s_1}J_1 \\
x_2 &\in T^{s_2}J_1
\end{align*}
Depending on whether $s_1 < \frac{r_1}{2}$ or $s_1 > \frac{r_1}{2}$, and similarly for $s_2$, the analysis splits into four cases.
- Case 1 ($s_1 < \frac{r_1}{2}$ and $s_2 < \frac{r_1}{2}$): In this case, we consider the average number of times $T^i x_1$ and $T^j x_2$ intersect $U$ as $i$ goes from $s_1$ to $r_1$, and $j$ goes from $s_2$ to $r_1$. Since both these orbits cover the entirety of $\left[\frac{r_1}{2}, r_1\right]$, the orbit averages must be really close, specifically within $\frac{3\varepsilon}{2}$ distance, but since $r_1$ is much larger than $t_1$ and $t_2$, the orbit averages must also be close to the space averages $\mu_1(U)$ and $\mu_2(U)$, which are at least $2\varepsilon$ distance, leading to a contradiction.
- Case 2 ($s_1 > \frac{r_1}{2}$ and $s_2 > \frac{r_1}{2}$): Dealing with this case is a matter of replacing $T$ with $T^{-1}$, which is where genericity with respect to $T^{-1}$ comes into play.
The other two cases are also dealt with similarly, by picking $T$ or $T^{-1}$ depending on whether $s_i < \frac{r_1}{2}$ or $s_i > \frac{r_1}{2}$.
This answers questions 1 and 3, but not 2. Also, there's a more general version of this result for systems with linear block growth (see Cyr and Kra's paper: https://arxiv.org/pdf/1505.02748.pdf).
