First question: UMP is, nomen es omen, most powerful. If both the sample size and the equivalence region are small, it may happen to the TOST that confidence intervals will hardly ever fit into the equivalence region. This results in nearly zero power. Also, the TOST is generally conservative (even with an $1-2\alpha$ confidence interval). Whenever the UMP exists, it will always have power $> \alpha$.
Second question: Sometimes an UMP doesn't exist. It is this strictly total positivity of order 3 that has to hold for the density, see the appendix of Wellek's textbook on equivalence and noninferiority tests. Intuitively, this condition guarantees that the power curve of the respective point hypothesis test has exactly one maximum. Then the critical values are the points where this power curve has level $\alpha$. That's why you find them with this $F_{1,n-1,\psi^2}$-distribution in this question: Obtaining $p$-values for UMP $t$ tests for equivalence.
Also if your equivalence hypothesis is not standardized, i.e. $\mu \in ]-\epsilon, \epsilon[$ instead of $\mu \in ]-\frac{\epsilon}{\sigma}, \frac{\epsilon}{\sigma}[$, then even for normally distributed data an UMP has a strange rejection area in the $(\hat{\mu},\hat{\sigma}^2)$-space. See Brown, Hwang and Munk (1997) as an example.
The most important is, as you mentioned, that confidence intervals on the observed scale are more instructive than $p$-values. So the ICH guidelines require confidence intervals. This leads automatically to the TOST, because if you supply a confidence interval to the $p$-value of the UMP, confidence interval and $p$-value may contradict each other. The UMP may be significant but the confidence interval still touches the hypothesis space. This is not desired.
In conclusion, if you use the equivalence test "internally", i.e. not for direct scientific reporting but only as part of some data mining algorithm e.g., UMP may be preferable if it exists. Otherwise take the TOST.
When conducting the Kolmogorov-Smirnov test, we assume $H_0:$ the two distributions are equivalent. We then calculate a test statistic and, if the corresponding $p$-value is small enough, we reject $H_0$ and conclude $H_A:$ the two distributions are different.
As far as hypothesis tests go, we use a $p$-value to quantify the amount of evidence we have to reject the null hypothesis. A $p$-value of 1 indicates that we have gathered no evidence to reject the null hypothesis. A $p$-value close to 0 indicates there is overwhelming evidence to reject the null hypothesis.
Let's assume we have data and calculate a $p$-value from the K-S test where $p=0.99.$ This indicates there is very little evidence to reject the null hypothesis. However, we cannot establish a standard of $\alpha=0.95$ such that $p>\alpha$ implies that we conclude the null hypothesis is correct. Further, I don't believe there is an alternate test that would allow us to conclude that the two distributions are the same.
What I believe you can do is to be entirely honest in the write-up or discussion. Mention that you ran a K-S test, report a $p$-value, and if the $p$-value is sufficiently high, then articulate that there is very little evidence to suggest that the two distributions are different. So, while you cannot conclude that the distributions are identical, you should be able to note that there is no evidence suggesting that the two distributions are different. As your sample size $n$ increases, the more faith you'll have in this answer.
It's not quite the answer that you were probably looking for, but it's not a total wash, either. Hope this helps!
Best Answer
The $1-2\alpha$ is not because you calculate the CI for each group separately. It is because you calculate the "inequivalence" to the upper and to the lower end separately. The parameter $\theta$ lies in the equivalence interval $[\epsilon_L, \epsilon_U]$ iff $$\theta \geq \epsilon_L \wedge \theta \leq \epsilon_U.$$
Each part is tested separately by a one sided test at level $1-\alpha$. Only if both tests are significant, we can conclude equivalence. (This is the very intuitive intersection-union-principle.) Turning this into a single confidence interval, we must remove $\alpha$ from both the upper and the lower probability mass of the CI. So we end up with $1-2\alpha$. The TOST-CI is simply the intersection of the one-sided CIs.
By the way, it is still possible to do the TOST with a $1-\alpha$ CI, but it would be unnecessarily conservative.