tl;dr: There are two commonly reported formulas for approximating $\text{var}(f(X))$, but one is notably better than the other. Since it isn't the "standard" Taylor expansion, where does it come from, and why is it better?
Details: Let $X$ be a real random variable and $f:\mathbb{R}\to\mathbb{R}$. There is a standard way to approximate $\text{var}(f(X))$ using Taylor expansions of moments, e.g. $E f(X)$, $E f^2(X)$, etc. Doing so yields the following second-order approximation: \begin{align} \operatorname{var}(f(X))\approx [f'(EX)]^2\operatorname{var}(X)-\frac{[f''(EX)]^2}{4}\operatorname{var}^2(X) := V_1. \end{align}
For a formal proof, see this post and answer. Oddly, this is not the formula quoted in the corresponding Wikipedia page: \begin{align} \operatorname{var}(f(X))\approx [f'(EX)]^2\operatorname{var}(X) + \frac{[f''(EX)]^2}{2}\operatorname{var}^2(X) := V_2. \end{align}
Note the difference in the coefficient of second term: $-1/4$ vs $+1/2$.
At first, I assumed this was a typo in the Wikipedia page. After running some quick simulations, however, it seems that the second approximation $V_2$ is much better than $V_1$! (Admittedly I did not run exhaustive tests but after a few dozen examples the difference was quite clear.)
In fact, here is a partial explanation for why $V_1$ often fails catastrophically: If $f'(EX)\approx 0$, then $V_1 < 0$. Presumably, this can be corrected by using a third-order approximation.
My question, though, is (a) How do we derive $V_2$, and (b) Why does it outperform $V_1$?