You're getting confused because the author you're reading has expressed Gauss' law using unusual and confusing notation, to the point where I would actually call it incorrect.
I can sort of see how the confusing equations could arise from a derivation of Gauss' law from Coulomb's law. From Coulomb's law, you have
$$\vec{E}(\vec{r}) = \frac{1}{4\pi\epsilon_0} \int \frac{\rho(\vec{r}')(\vec{r}-\vec{r}')}{|\vec{r}-\vec{r}'|^3} \, d^3 \vec{r}'\ .$$
Taking the divergence of both sides with respect to $\vec{r}$, and taking advantage of the identity
$$\nabla \cdot \left(\frac{\vec{r}}{|\vec{r}|^3}\right) = 4\pi \delta(\vec{r})$$
along the way, you end up with
$$\nabla\cdot\vec{E}(\vec{r}) = \frac{1}{\epsilon_0} \int \rho(\vec{r}')\ \delta(\vec{r}-\vec{r}')\, d^3 \vec{r}'\ .$$
But this is where I disagree with how your author has expressed things. $\delta(\vec{r}-\vec{r}')$ is zero everywhere except where $\vec{r}=\vec{r}'$. The normal way to express $\delta$'s sifting property on that last integral would be to say that the integral evaluates to $\rho(\vec{r})$, not to say that the integral evaluates to $\rho(\vec{r}')$. If you say that the integral evaluates to $\rho(\vec{r}')$, you'd have to add that the only $\vec{r}'$ involved is where $\vec{r}'=\vec{r}$, in which case it make more sense to just say $\rho(\vec{r})$ instead of $\rho(\vec{r}')$. And that would indeed be what is normally done; Gauss' law would more commonly be expressed as
$$\nabla\cdot\vec{E}(\vec{r})=\frac{\rho(\vec{r})}{\epsilon_0}\ .$$