Conditional variance of the absolute sum of zero-mean i.i.d. random variables Why does the following conditional variance formula hold, and does it hold in general?
$$\text{Var} \left( \left \vert \sum^n_{i=1} U_i \right \vert \, \middle | \,D_n \right) = n \text{Var}(U_i \vert D_n).$$
Here $U_i$ are zero-mean independent and identically distributed random variables, and nothing more about their distribution is stated. Further information on $U_i$ and $D_n$ is provided below.
I am comfortable with the standard result that for i.i.d. random variables $X_i$,
$$\text{Var} \left( \sum^n_{i=1} X_i \right) = \sum^n_{i=1} \text{Var}(X_i) = n \text{Var}(X_i),$$
but the presence of the absolute values in the formula of interest has thrown me.

Further information on $U_i$ and $D_n$.
In response to the helpful comments and answer by James Martin and Ben, it is clear that the equation does not hold in general, and that additional context on the $U_i$ and $D_n$ is required. Here is an extract of the proof where the equation is used:




Where $D_n = {(X_1, Y_1,), \dots, (X_n, Y_n)}$ and $D_n' = (X_1', Y_1'), \dots (X_n', Y_n')$ are two sets of i.i.d. random variables sampled from unknown joint distribution $D_n, D_n' \overset{i.i.d}{\sim} P(X, Y)$. The conditional expectation in $U_i$ is such that $\mathbb{E}_{D_n' | D_n}[\cdot \vert D_n]$. $f: \mathbb{R}^d \rightarrow \{0, 1 \}$ is a classifier function , while $\tilde{f}$ is a random classifier function, dependent on $D_n$.
The argument comes from the proof of the Vapnik-Chervonenkis inequality in an expository note by Robert Nowak (2009), following the strategy of Devroye,  Györfi and Lugosi (1996) in their proof of Glivenko-Cantelli theorem.

$\newcommand{\Var}{\operatorname{Var}}$
In response to the edited version.
I think there's a glitch in that the "$=$" sign ought to be a "$\geq$" sign. With that change, everything seems to be (at least locally) OK.
A general fact is that $\Var(|X|)\leq \Var(X)$ for any random variable $X$. (Since $\Var(X)-\Var(|X|)=E(|X|)^2 - E(X)^2$,
and $|E(X)|\leq E(|X|)$.)
So (applying the above conditional on $D_n$) we get that $\Var\left(|\sum_{i=1}^n U_i| \big| D_n\right)
\leq
\Var\left(\sum_{i=1}^n U_i \big| D_n\right)$.
Now we use the asserted fact (first line of the excerpt you quoted) that, conditional on $D_n$, the $U_i$ are i.i.d. (N.B. it's important that the $U_i$ are i.i.d. conditional on $D_n$; this is a different statement from saying that that $U_i$ are unconditionally i.i.d.)
That gives us that
$\Var\left(\sum_{i=1}^n U_i \big| D_n\right)
=
\sum_{i=1}^n\Var\left( U_i \big| D_n\right)
=
n\Var\left(U_1\big|D_n\right)$.
Finally the next line is using the fact that $\Var(U_1\big| D_n)\leq 1/4$ (because conditional on $D_n$, $U_1$ is a shift of a Bernoulli random variable, and any Bernoulli random variable has variance at most $1/4$).
