Let's take the case of the Poisson RV's as an example of what is going on.
The distribution of a Poisson RV is
$$
P_\lambda(k)=\frac{\lambda^ke^{-\lambda}}{k!}\tag1
$$
where $\lambda$ is the expected value of the RV; it is also the variance of the RV. Summing $n$ RVs gives a distribution of
$$
P_{n\lambda}(k)=\frac{(n\lambda)^ke^{-n\lambda}}{k!}\tag2
$$
To make the mean $0$, we translate by $n\lambda$:
$$
P_{n\lambda}(k+n\lambda)=\frac{(n\lambda)^{k+n\lambda}e^{-n\lambda}}{(k+n\lambda)!}\tag3
$$
Note that as with any convolution of random variables, this produces a thinned out distribution; that is, the distribution gets spread out over a larger range and the probability for any given range is decreased commensurately. To counter the effect of this thinning, we scale back by $\sqrt{n}$ to get a distribution with mean $0$ and variance $\lambda$:
$$
\begin{align}
\sqrt{n}P_{n\lambda}(\sqrt{n}k+n\lambda)
&=\sqrt{n}\frac{(n\lambda)^{\sqrt{n}k+n\lambda}e^{-n\lambda}}{\left(\sqrt{n}k+n\lambda\right)!}\tag4\\[6pt]
&\sim\frac{\sqrt{n}}{\sqrt{2\pi}}\frac{(n\lambda)^{\sqrt{n}k+n\lambda}e^{-n\lambda}e^{\sqrt{n}k+n\lambda}}{\left(\sqrt{n}k+n\lambda\right)^{\sqrt{n}k+n\lambda+1/2}}\tag5\\[3pt]
&=\frac1{\sqrt{2\pi\lambda}}\frac{(n\lambda)^{\sqrt{n}k+n\lambda+1/2}e^{\sqrt{n}k}}{\left(\sqrt{n}k+n\lambda\right)^{\sqrt{n}k+n\lambda+1/2}}\tag6\\[3pt]
&=\frac1{\sqrt{2\pi\lambda}}\frac{e^{\sqrt{n}k}}{\left(\vcenter{1+\frac{k}{\sqrt{n}\lambda}}\right)^{\sqrt{n}k+n\lambda+1/2}}\tag7\\
&\sim\frac1{\sqrt{2\pi\lambda}}e^{-\frac{k^2}{2\lambda}}\tag8
\end{align}
$$
Explanation:
$(4)$: substitute $\sqrt{n}k\mapsto k$ in $(3)$ and multiply by $\sqrt{n}$
$(5)$: apply Stirling's Approximation
$(6)$: cancel $e^{n\lambda}$ and $(n\lambda)^{1/2}$
$(7)$: cancel $(n\lambda)^{\sqrt{n}k+n\lambda+1/2}$
$(8)$: $\left(\vcenter{1+\frac{k}{\sqrt{n}\lambda}}\right)^{\sqrt{n}k}\sim e^{\frac{k^2}\lambda}$
$\phantom{\text{(8):}}$ $\left(\vcenter{1+\frac{k}{\sqrt{n}\lambda}}\right)^{n\lambda}\sim e^{\sqrt{n}k-\frac{k^2}{2\lambda}}$
$\phantom{\text{(8):}}$ $\left(\vcenter{1+\frac{k}{\sqrt{n}\lambda}}\right)^{1/2}\sim 1$
Formula $(8)$ is normal distribution with mean $0$ and variance $\lambda$.
Here is the Poisson distribution, $P_{n\lambda}$ for $\lambda=1$, scaled and plotted for various $n$. The points are placed so that $\sqrt{n}k+n\lambda\in\mathbb{Z}$.
The continuous curve is the Normal distribution with variance $1$.
Thus, the Poisson distribution, when scaled to counter the thinning, tends to a normal distribution.
At no point is the Poisson distribution a normal distribution. For one thing for any $n$, $P_{n\lambda}$ is a discrete distribution; the normal distribution is a continuous distribution. This is a property of limits: it is not necessary that at any point, a sequence equals its limit.
