Question 1 indeed seems right until one reads your comment. It seems right, because
we are thinking that $D_U$ is the branch divisor. It is indeed right if you take
$D_U=D_{m,U}$ in place of your choice.

On the other hand, as stated, Question 1 could not be correct!! Observe that the right
hand side is independent of $m$ while the left hand side can get bigger and bigger
as $m$ grows.

The crux of the matter is that $D_{V_m}$ should be compared to $D_{m,U}$ and not
$D_U$.

Here is some evidence: Observe that $B=D_{m,U}$ is the branch divisor of $\pi_m$. Let
$R=\frac 1m\pi^*B$ (a $\mathbb Q$-divisor if you want) be the $m^\text{th}$ root of
$B$. Then we have the (logarithmic) ramification formula: $$K_{V_m}+ R \sim \pi^*(K_U
+ B).\tag{$*$}$$

Next observe that $B\sim mD_U$ and hence $D_U$ cannot contain $B$ (which is assumed
to be irreducible), so $D_{V_m}=\pi^*D_U$ and then $D_{V_m}\sim_{\mathbb Q}R$.

So, we obtain by ($*$) that $$\det\Omega_{V_m}(\log D_{V_m})\sim_{\mathbb Q}\det
\Omega_{V_m}(\log R)\sim \pi^*(\det\Omega_U(\log B)),\tag{+}$$
which is strictly bigger than
$\pi^*(\det\Omega_U(\log D_U))$.

**Remarks**

1) It is entirely unimportant that the linear system comes from $-K_U$. This was not
used anywhere, but in the last sentence one may notice that with that definition
$\det\Omega_U(\log D_U)\simeq \mathscr O_U$, which might make it easier to believe
that the claim and proof here is correct.

2) If one defines $D_U=D_{m,U}$, then Question 1 is correct, this is an easy local
calculation.

3) Under #2 above Question 1 implies Question 2. The proof is, as Donu suggests,
simply using the projection formula and then simply applying the same formula for the
push-forward of the structure sheaf for a cyclic cover.

**Addition:**
The recent edit to your question seems to indicate that perhaps you still expect that Question 2 might be true even if Question 1 is not. I don't think that's going to happen either. (I thought this would be kind of clear from the above).

By the above computation and remarks one can see that
$$\pi_*\Omega_{V_m}(\log R)\sim \Omega_U(\log B)\otimes \pi_*\mathscr O_{V_m}\sim \Omega_U(\log B)\otimes (\oplus_j\mathscr O_{U}(-jB)),$$
so the $\zeta$-eigensheaf for this pushforward is
$\Omega_U(\log B)(-B)$.

Now since $\det\Omega_{V_m}(\log D_{V_m})\sim_{\mathbb Q}\det
\Omega_{V_m}(\log R)$, counting degrees should show you that the answer to Question 2 is also negative, unless $\dim U=1$. On the other hand, if $\dim U=1$, you will be hard pressed to find a $D_{m,U}$ that is both smooth and irreducible for $m>1$.