The slides you link to are somewhat confusing, leaving out steps and making a few typos, but they are ultimately correct. It will help to answer question 2 first, then 1, and then finally derive the symmetrizing transformation $A(u) = \int_{-\infty}^u \frac{1}{[V(\theta)]^{1/3}} d\theta$.
Question 2. We are analyzing $\bar{X}$ as it the mean of a sample of size $N$ of i.i.d. random variables $X_1, ..., X_N$. This is an important quantity because sampling the same distribution and taking the mean happens all the time in science. We want to know how close $\bar{X}$ is to the true mean $\mu$. The Central Limit Theorem says it will converge to $\mu$ as $N \to \infty$ but we would like to know the variance and skewness of $\bar{X}$.
Question 1. Your Taylor series approximation is not incorrect, but we need to be careful about keeping track of $\bar{X}$ vs. $X_i$ and powers of $N$ to get to the same conclusion as the slides. We'll start with the definitions of $\bar{X}$ and central moments of $X_i$ and derive the formula for $\kappa_3(h(\bar{X}))$:
$\bar{X} = \frac{1}{N}\sum_{i=1}^N X_i$
$\mathbb{E}[X_i] = \mu$
$V(X_i) = \mathbb{E}[(X_i - \mu)^2] = \sigma^2$
$\kappa_3(X_i) = \mathbb{E}[(X_i - \mu)^3]$
Now, the central moments of $\bar{X}$:
$\mathbb{E}[\bar{X}] = \frac{1}{N}\sum_{i=1}^N \mathbb{E}[X_i] = \frac{1}{N}(N\mu) = \mu$
$\begin{align}
V(\bar{X}) &=\mathbb{E}[(\bar{X} - \mu)^2]\\
&=\mathbb{E}[\Big((\frac{1}{N}\sum_{i=1}^N X_i) - \mu\Big)^2]\\
&=\mathbb{E}[\Big(\frac{1}{N}\sum_{i=1}^N (X_i - \mu)\Big)^2]\\
&=\frac{1}{N^2}\Big(N\mathbb{E}[(X_i - \mu)^2] + N(N-1)\mathbb{E}[X_i - \mu]\mathbb{E}[X_j - \mu]\Big)\\
&= \frac{1}{N}\sigma^2
\end{align}$
The last step follows since $\mathbb{E}[X_i - \mu] = 0$, and $\mathbb{E}[(X_i - \mu)^2] = \sigma^2$. This might not have been the easiest derivation of $V(\bar{X})$, but it is the same process we need to do to find $\kappa_3(\bar{X})$ and $\kappa_3(h(\bar{X}))$, where we break up a product of a summation and count the number of terms with powers of different variables. In the above case, there were $N$ terms that were of the form $(X_i - \mu)^2$ and $N(N-1)$ terms of the form $(X_i - \mu)(X_j - \mu)$.
$\begin{align}
\kappa_3(\bar{X}) &= \mathbb{E}[(\bar{X}-\mu)^3)]\\
&= \mathbb{E}[\Big((\frac{1}{N}\sum_{i=1}^N X_i) - \mu\Big)^3]\\
&= \mathbb{E}[\Big(\frac{1}{N}\sum_{i=1}^N (X_i - \mu)\Big)^3]\\
&= \frac{1}{N^3}\Big(N\mathbb{E}[(X_i - \mu)^3] + 3N(N-1)\mathbb{E}[(X_i - \mu)\mathbb{E}[(X_j - \mu)^2]+N(N-1)(N-2)\mathbb{E}[(X_i - \mu)]\mathbb{E}[(X_j - \mu)]\mathbb{E}[(X_k - \mu)]\\
&= \frac{1}{N^2}\mathbb{E}[(X_i - \mu)^3]\\
&= \frac{\kappa_3(X_i)}{N^2}
\end{align}$
Next, we will expand $h(\bar{X})$ in a Taylor series as you have:
$h(\bar{X}) = h(\mu) + h'(\mu)(\bar{X} - \mu) + \frac{1}{2}h''(\mu)(\bar{X}-\mu)^2 + \frac{1}{3}h'''(\mu)(\bar{X}-\mu)^3 + ...$
$\begin{align}
\mathbb{E}[h(\bar{X})] &= h(\mu) + h'(\mu)\mathbb{E}[\bar{X} - \mu] + \frac{1}{2}h''(\mu)\mathbb{E}[(\bar{X}-\mu)^2] + \frac{1}{3}h'''(\mu)\mathbb{E}[(\bar{X}-\mu)^3] + ...\\
&= h(\mu) + \frac{1}{2}h''(\mu)\frac{\sigma^2}{N} + \frac{1}{3}h'''(\mu)\frac{\kappa_3(X_i)}{N^2} + ...\\
\end{align}$
With some more effort you could prove the rest of the terms are $O(N^{-3})$. Finally, since $\kappa_3(h(\bar{X})) = \mathbb{E}[(h(\bar{X})-\mathbb{E}[h(\bar{X})])^3]$, (which is not the same as $\mathbb{E}[(h(\bar{X})-h(\mu))^3]$), we again make a similar computation:
$\begin{align}
\kappa_3(h(\bar{X})) &= \mathbb{E}[(h(\bar{X})-\mathbb{E}[h(\bar{X})])^3]\\
&=\mathbb{E}\Big[\Big(h(\mu) + h'(\mu)(\bar{X} - \mu) + \frac{1}{2}h''(\mu)(\bar{X}-\mu)^2 + O((\bar{X}-\mu)^3) - h(\mu) - \frac{1}{2}h''(\mu)\frac{\sigma^2}{N} - O(N^{-2})\Big)^3\Big]
\end{align}$
We are only interested in the terms resulting in order $O(N^{-2})$, and with extra work you could show that you do not need the terms "$O((\bar{X}-\mu)^3)$" or "$- O(N^{-2})$" before taking the third power, as they will only result in terms of order $O(N^{-3})$. So, simplifying, we get
$\begin{align}
\kappa_3(h(\bar{X})) &= \mathbb{E}\Big[\Big(h'(\mu)(\bar{X} - \mu) + \frac{1}{2}h''(\mu)(\bar{X}-\mu)^2 - \frac{1}{2}h''(\mu)\frac{\sigma^2}{N})\Big)^3\Big]\\
&=\mathbb{E}\Big[h'(\mu)^3(\bar{X} - \mu)^3 + \frac{1}{8}h''(\mu)^3(\bar{X}-\mu)^6 - \frac{1}{8}h''(\mu)^3\frac{\sigma^6}{N^3} + \frac{3}{2}h'(\mu)^2h''(\mu)(\bar{X}-\mu)^4 + \frac{3}{4}h'(\mu)h''(\mu)(\bar{X}-\mu)^5 - \frac{3}{2}h'(\mu)^2h''(\mu)(\bar{X} - \mu)^2\frac{\sigma^2}{N} + O(N^{-3})\Big]
\end{align}$
I left off some terms that were obviously $O(N^{-3})$ in this product. You'll have to convince yourself that the terms $\mathbb{E}[(\bar{X}-\mu)^5]$ and $\mathbb{E}[(\bar{X}-\mu)^6]$ are $O(N^{-3})$ as well. However,
$\begin{align}
\mathbb{E}[(\bar{X}-\mu)^4] &= \mathbb{E}[\frac{1}{N^4}\Big(\sum_{i=1}^N(\bar{X}-\mu)\Big)^4]\\
&=\frac{1}{N^4}\Big(N\mathbb{E}[(X_i-\mu)^4] + 3N(N-1)\mathbb{E}[(X_i-\mu)^2]\mathbb{E}[(X_j-\mu)^2] + 0\Big)\\
&=\frac{3}{N^2}\sigma^4 + O(N^{-3})
\end{align}$
Then distributing the expectation on our equation for $\kappa_3(h(\bar{X}))$, we have
$\begin{align}\kappa_3(h(\bar{X})) &= h'(\mu)^3\mathbb{E}[(\bar{X} - \mu)^3] + \frac{3}{2}h'(\mu)^2h''(\mu)\mathbb{E}[(\bar{X}-\mu)^4] - \frac{3}{2}h'(\mu)^2h''(\mu)\mathbb{E}[(\bar{X} - \mu)^2]\frac{\sigma^2}{N} + O(N^{-3})\\
&= h'(\mu)^3\frac{\kappa_3(X_i)}{N^2} + \frac{9}{2}h'(\mu)^2h''(\mu)\frac{\sigma^4}{N^2} - \frac{3}{2}h'(\mu)^2h''(\mu)\frac{\sigma^4}{N^2} + O(N^{-3})\\
&=h'(\mu)^3\frac{\kappa_3(X_i)}{N^2} + 3h'(\mu)^2h''(\mu)\frac{\sigma^4}{N^2} + O(N^{-3})
\end{align}$
This concludes the derivation of $\kappa_3(h(\bar{X}))$. Now, at last, we will derive the symmetrizing transform $A(u) = \int_{-\infty}^u \frac{1}{[V(\theta)]^{1/3}} d\theta$.
For this transformation, it is important that $X_i$ is from an exponential family distribution, and in particular a natural exponential family (or it has been transformed into this distribution), of the form $f_{X_i}(x;\theta) = h(x)\exp(\theta x - b(\theta))$
In this case, the cumulants of the distribution are given by $\kappa_k = b^{(k)}(\theta)$. So $\mu = b'(\theta)$, $\sigma^2 = V(\theta) = b''(\theta)$, and $\kappa_3 = b'''(\theta)$. We can write the parameter $\theta$ as a function of $\mu$ just taking the inverse of $b'$, writing $\theta(\mu) = (b')^{-1}(\mu)$. Then
$\theta'(\mu) = \frac{1}{b''((b')^{-1}(\mu))} = \frac{1}{b''(\theta))} = \frac{1}{\sigma^2}$
Next we can write the variance as a function of $\mu$, and call this function $\bar{V}$:
$\bar{V}(\mu) = V(\theta(\mu)) = b''(\theta(\mu))$
Then
$\frac{d}{d\mu}\bar{V}(\mu) = V'(\theta(\mu))\theta'(\mu) = b'''(\theta)\frac{1}{\sigma^2} = \frac{\kappa_3}{\sigma^2}$
So as a function of $\mu$, $\kappa_3(\mu) = \bar{V}'(\mu)\bar{V}(\mu)$.
Now, for the symmetrizing transformation, we want to reduce the skewness of $h(\bar{X})$ by making $h'(\mu)^3\frac{\kappa_3(X_i)}{N^2} + 3h'(\mu)^2h''(\mu)\frac{\sigma^4}{N^2} = 0$ so that $h(\bar{X})$ is $O(N^{-3})$. Thus, we want
$h'(\mu)^3\kappa_3(X_i) + 3h'(\mu)^2h''(\mu)\sigma^4 = 0$
Substituting our expressions for $\sigma^2$ and $\kappa_3$ as functions of $\mu$, we have:
$h'(\mu)^3\bar{V}'(\mu)\bar{V}(\mu) + 3h'(\mu)^2h''(\mu)\bar{V}(\mu)^2 = 0$
So $h'(\mu)^3\bar{V}'(\mu) + 3h'(\mu)^2h''(\mu)\bar{V}(\mu) = 0$, leading to $\frac{d}{d\mu}(h'(\mu)^3\bar{V}(\mu)) = 0$.
One solution to this differential equation is:
$h'(\mu)^3\bar{V}(\mu) = 1$,
$h'(\mu) = \frac{1}{[\bar{V}(\mu)]^{1/3}}$
So, $h(\mu) = \int_c^\mu \frac{1}{[\bar{V}(\theta)]^{1/3}} d\theta$, for any constant, $c$. This gives us the symmetrizing transformation $A(u) = \int_{-\infty}^u \frac{1}{[V(\theta)]^{1/3}} d\theta$, where $V$ is the variance as a function of the mean in a natural exponential family.
