Let $|\vec{r}| \gg\left|\vec{r^{\prime}}\right|$ and $x:=\frac{\vec{r}^{2}-2 \vec{r^{2}} \cdot \vec{r^{\prime}}}{\vec{r}^{2}} $ be a small number.
$$\frac{1}{\left|\vec{r}-\vec{r^{\prime}}\right|}=\frac{1}{\sqrt{\vec{r}^{2}+{\vec{r^{\prime}}}^{2}-2 \vec{r} \cdot \vec{r^{\prime}}}}=\frac{1}{|\vec{r}| \sqrt{1+\frac{{\vec{r^{\prime}}}^{2}-2 \vec{r} \cdot \vec{r^{\prime}}}{\vec{r}^{2}}} }=\frac{1}{|\vec{r}| \sqrt{1+x} } $$
With the binomial Series:
$$ \frac{1}{\sqrt{1+x}}=(1+x)^{-\frac{1}{2}}=1-\frac{1}{2} x+\frac{3}{8} x^{2}-\ldots$$
we get in first order:
$$ \frac{1}{\left|\vec{r}-\vec{r^{\prime}}\right|}=\frac{1}{|\vec{r}|}+\frac{\vec{r} \cdot \vec{r^{\prime}}}{|\vec{r}|^{3}} $$
I don't understand the last step. We use the Taylor series of the binomial equation, but at what point? Can someone explain the last step in more detail?
It seems like you're happy with the equality:
$$ \frac{1}{\lVert \vec{r} - \vec{r'} \rVert} = \frac{1}{ \lVert \vec{r} \rVert \sqrt{1+x}} $$
where $x = \frac{\vec{r'}^2 - 2 \vec{r} \cdot \vec{r'}}{\vec{r}^2}$ (notice there's a typo in the first line of your question, but during the manipulation in your second line you correct it).
Using the fact that $$\frac{1}{\sqrt{1+x}} = (1+x)^{-1/2} = 1 - \frac{1}{2} x + \frac{3}{8} x^2 + \ldots$$
we see (keeping only first order terms)
$$ \frac{1}{ \lVert \vec{r} \rVert} \frac{1}{\sqrt{1+x}} = \frac{1}{\lVert \vec{r} \rVert} \left ( 1 - \frac{1}{2} x + O(x^2) \right ) \approx \frac{1}{\lVert \vec{r} \rVert} - \frac{1}{2} \frac{x}{\lVert \vec{r} \rVert} $$
Now substituting our definition for $x$ back in, we see this is all equal to
$$ \frac{1}{\lVert \vec{r} \rVert} - \frac{1}{2} \frac{\vec{r'}^2 - 2 \vec{r} \cdot \vec{r'}}{\vec{r}^2 \lVert \vec{r} \rVert} $$
Remembering that $\vec{r}^2 = \vec{r} \cdot \vec{r} = \lVert \vec{r} \lVert^2$, this becomes
$$ \frac{1}{\lVert \vec{r} \rVert} - \frac{1}{2} \left ( \frac{\lVert \vec{r'} \rVert^2}{\lVert \vec{r} \rVert^3} + \frac{-2 \vec{r} \cdot \vec{r'}}{\lVert \vec{r} \rVert^3} \right ) $$
But since $\lVert r' \rVert \ll \lVert r \rVert$, the $\frac{\lVert \vec{r'} \rVert^2}{\lVert \vec{r} \rVert^3}$ term is small (so isn't seen in the first order terms), ignoring it we're left with
$$ \frac{1}{\lVert \vec{r} \rVert} - \frac{1}{2} \frac{-2 \vec{r} \cdot \vec{r'}}{\lVert \vec{r} \rVert^3} = \frac{1}{\lVert \vec{r} \rVert} + \frac{\vec{r} \cdot \vec{r'}}{\lVert \vec{r} \rVert^3} $$
Comparing with where we started, we see (to first order)
$$ \frac{1}{\lVert \vec{r} - \vec{r'} \rVert} \approx \frac{1}{\lVert \vec{r} \rVert} + \frac{\vec{r} \cdot \vec{r'}}{\lVert \vec{r} \rVert^3} $$
as claimed.
I hope this helps ^_^