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.

Your equation doesn't hold in general.  As a simple counter-example, consider random variables that are conditionally IID with distribution $\mathbb{P}(U_i=-1|D_n) = \mathbb{P}(U_i=1|D_n)=\tfrac{1}{2}$.$^\dagger$  In this simple case you get:
$$\begin{align}
\mathbb{E}(U_1 + U_2 | D_n) 
&= (-2) \cdot \frac{1}{4} + 0 \cdot \frac{1}{2} + 2 \cdot \frac{1}{4} = 0, \\[12pt]
\mathbb{V}(U_1 + U_2 | D_n) 
&= (-2)^2 \cdot \frac{1}{4} + 0^2 \cdot \frac{1}{2} + 2^2 \cdot \frac{1}{4} = 2, \\[12pt]
\mathbb{E}(|U_1 + U_2| | D_n) 
&= |-2| \cdot \frac{1}{4} + |0| \cdot \frac{1}{2} + |2| \cdot \frac{1}{4} = 1, \\[12pt]
\mathbb{V}(|U_1 + U_2| | D_n) 
&= (|-2|-1)^2 \cdot \frac{1}{4} + (|0|-1)^2 \cdot \frac{1}{2} + (|2|-1)^2 \cdot \frac{1}{4} = 1, \\[12pt]
\end{align}$$
so you have:
$$\mathbb{V}(|U_1 + U_2| | D_n) = 1 \neq 2 = \mathbb{V}(U_1 + U_2 | D_n).$$
In fact, as a general rule (stated here unconditionally but it also holds conditionally), you get:
$$\begin{align}
\mathbb{V} \Bigg( \bigg| \sum_{i=1}^n U_i \bigg| \Bigg) 
&= \mathbb{E} \Bigg( \bigg| \sum_{i=1}^n U_i \bigg| ^2 \Bigg) - \mathbb{E} \Bigg( \bigg| \sum_{i=1}^n U_i \bigg| \Bigg)^2  \\[6pt]
&\leqslant \mathbb{E} \Bigg( \bigg| \sum_{i=1}^n U_i \bigg| ^2 \Bigg) - \mathbb{E} \Bigg( \sum_{i=1}^n U_i \Bigg)^2  \\[6pt]
&= \mathbb{E} \Bigg( \bigg( \sum_{i=1}^n U_i \bigg) ^2 \Bigg) - \mathbb{E} \Bigg( \sum_{i=1}^n U_i \Bigg)^2  \\[6pt]
&= \mathbb{V} \Bigg( \sum_{i=1}^n U_i \Bigg) \\[6pt]
&= \sum_{i=1}^n \mathbb{V}(U_i) + \sum_{i \neq j} \mathbb{C}(U_i,U_j). \\[6pt]
\end{align}$$
In regard to your updated information, the result still does not hold with the additional information.  (And indeed, the additional information shows that the situation is just a scaled version of what I have described above, so the result is generally false in that case.)  It appears to me that this is probably just an error in the proof, using an invalid step.  Substiting the correct inequality in the erroneous step still gives a valid proof, so the proof itself still works once corrected.

$^\dagger$ You haven't given us any information on what $D_n$ is, so it is essentially irrelevant here.  Since you haven't told us anything about $D_n$ you can just posit that the values for $U_i$ are marginally IID and that $D_n$ is some irrelevant variable that is independent of these.
