If $X \sim \text{N}(0, \sigma^2)$ it is well-known that we have the second-order Taylor approximation:
$$\mathbb{V}[f(X)] \approx f'(\mu)^2 \cdot \sigma^2 + \frac{f''(\mu)^2}{2} \cdot \sigma^4.$$
What is the general form for this variance approximation when we have a known underlying distribution, but we do not assume normality or zero mean?
The general form of the variance approximation depends on the first four moments of the distribution. To facilitate our analysis, we suppose that $X$ has mean $\mu$, variance $\sigma^2$, skewness $\gamma$ and kurtosis $\kappa$, and assume that the kurtosis is finite. The general variance approximation is:
$$\boxed{\begin{equation} \begin{aligned} \mathbb{V}[f(X)] &\approx ( f''(\mu)^2 \mu^2 - f'(\mu)f''(\mu) \mu + f'(\mu)^2 ) \cdot \sigma^2 \\[6pt] &\quad - \frac{f''(\mu)(f'(\mu) + \mu f''(\mu))}{2} \cdot \gamma \sigma^3 + \frac{f''(\mu)^2}{4} \cdot (\kappa-1) \sigma^4. \\[6pt] \end{aligned} \end{equation}}$$
The special case shown in the question is for a centred normal distribution, where we have $\mu=0$, $\gamma=0$ and $\kappa=3$.
Working for this result: Let's start by taking the second-order Taylor expansion of $f$ around the mean $\mu$. This gives us the approximating quadratic function:
$$\begin{equation} \begin{aligned} \hat{f}(x) &= f(\mu) + f'(\mu) \cdot (x-\mu) + \frac{1}{2} \cdot f''(\mu) \cdot (x-\mu)^2 \\[6pt] &= \frac{f''(\mu)}{2} \cdot x^2 + (f'(\mu) - \mu f''(\mu)) \cdot x + \text{const}. \\[6pt] \end{aligned} \end{equation}$$
The variance of this quadratic function is:
$$\begin{equation} \begin{aligned} \mathbb{V}[\hat{f}(X)] &= \mathbb{V} \Big[ \frac{f''(\mu)}{2} \cdot X^2 + (f'(\mu) - a f''(\mu)) \cdot X \Big] \\[6pt] &= \frac{f''(\mu)^2}{4} \cdot \mathbb{V}[X^2] + (f'(\mu) - \mu f''(\mu))^2 \cdot \mathbb{V}[X] \\[6pt] &\quad + \frac{f''(\mu)}{2} \cdot (f'(\mu) - \mu f''(\mu)) \cdot \mathbb{C}[X,X^2]. \\[6pt] \end{aligned} \end{equation}$$
Using the relationships between the moments and cumulants, it can be shown (see here and here for derivation) that:
$$\begin{equation} \begin{aligned} \mathbb{V}[X] &= \sigma^2, \\[6pt] \mathbb{V}[X^2] &= 4 \mu^2 \sigma^2 - 4 \mu \gamma \sigma^3 + (\kappa-1) \sigma^4, \\[6pt] \mathbb{C}[X,X^2] &= 2 \mu \sigma^2 + \gamma \sigma^3. \\[6pt] \end{aligned} \end{equation}$$
We therefore have:
$$\begin{equation} \begin{aligned} \mathbb{V}[\hat{f}(X)] &= \frac{f''(\mu)^2}{4} \cdot \Big[ 4 \mu^2 \sigma^2 - 4 \mu \gamma \sigma^3 + (\kappa-1) \sigma^4 \Big] + (f'(\mu) - \mu f''(\mu))^2 \cdot \sigma^2 \\[6pt] &\quad + \frac{f''(\mu)}{2} \cdot (f'(\mu) - \mu f''(\mu)) \cdot \Big[ 2 \mu \sigma^2 + \gamma \sigma^3 \Big] \\[6pt] &= f''(\mu)^2 \mu^2 \sigma^2 + (f'(\mu) - \mu f''(\mu))^2 \sigma^2 + (f'(\mu) - \mu f''(\mu)) f''(\mu) \mu \sigma^2 \\[6pt] &\quad - f''(\mu)^2 \mu \gamma \sigma^3 + (f'(\mu) - \mu f''(\mu)) \frac{f''(\mu)}{2} \gamma \sigma^3 \\[6pt] &\quad + \frac{f''(\mu)^2}{4} (\kappa-1) \sigma^4 \\[6pt] &= \Big[ f''(\mu)^2 \mu^2 + (f'(\mu) - \mu f''(\mu)) f''(\mu) \mu + (f'(\mu) - \mu f''(\mu))^2 \Big] \sigma^2 \\[6pt] &\quad - \Big[ f''(\mu)^2 \mu + (f'(\mu) - \mu f''(\mu)) \frac{f''(\mu)}{2} \Big] \gamma \sigma^3 \\[6pt] &\quad + \frac{f''(\mu)^2}{4} (\kappa-1) \sigma^4 \\[6pt] &= \Big[ f'(\mu) f''(\mu) \mu + (f'(\mu) - \mu f''(\mu))^2 \Big] \sigma^2 \\[6pt] &\quad - \Big[ f''(\mu)^2 \mu + (f'(\mu) - \mu f''(\mu)) \frac{f''(\mu)}{2} \Big] \gamma \sigma^3 \\[6pt] &\quad + \frac{f''(\mu)^2}{4} (\kappa-1) \sigma^4. \\[6pt] \end{aligned} \end{equation}$$
This gives us the true variance of the second-order Taylor approximation to the function. Observe that the approximation depends on the skewness and kurtosis of the distribution, not just its mean and variance. Now, using this variance as the approximation to the variance of the actual function (i.e., taking $\mathbb{V}[f(X)] \approx \mathbb{V}[\hat{f}(X)]$) gives the result shown above.