In the attached picture, the author evaluated first $ \frac{∂}{∂x_{j}} \phi(\vec{r})$ then to evaluate, in the line below it, $\frac{∂}{∂x_{j}} \phi(0)$, he simply substituted $\vec{r}=0$ in the calculation above it, so can that be done? another question is had we started with equation (1.29) shouldn't $\frac{∂}{∂x_{j}} \phi(0)$ be zero to begin with?
What $\mathbf{r} = 0$ means in this context $\mathbf{r} = (0,\dots 0)$ so with that in mind $$ \vert\mathbf{r} - \mathbf{r_0}\vert = \vert- \mathbf{r_0}\vert = r_0 $$
This is evaluating the potential at a single point, so is a constant as it is now independent of $\mathbf{r}$, so the potential derivative would indeed be $0$ evaluated at $\mathbf{r} = 0$. Sidenote, the derivative would be $0$ at any specific point e.g. $\mathbf{r} = \mathbf{r_{1}}$
To answer your follow up question - it appears the derivation is really computing $$ \frac{\partial \phi(\mathbf{r})}{\partial x_j}(\mathbf{r} = 0) = \frac{\partial \phi(\mathbf{r})}{\partial x_j}\big\vert_{\mathbf{r} =0} = -\frac{\alpha}{r_0^2}\frac{-x_{j0}}{r_0} = \frac{\alpha}{r_0^3}x_{j0} $$
