Don't let the vector notation confuse you - it's the same situation as an ordinary multidimensional Taylor expansion. After all, in each case, the end result is a scalar, not a vector. If you split each vector into its components (for example, $\vec{r}=x\hat{x}+y\hat{y}+z\hat{z}$, $\vec{k}=k_x\hat{x}+k_y\hat{y}+k_z\hat{z}$, and $\vec{r_0}=x_0\hat{x}+y_0\hat{y}+z_0\hat{z}$) and write the expressions out that way, it might be clearer:
$$\psi(\vec{r})\to\psi(x,y,z)=\left(\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}\right)^{3/2}$$
$$\psi(\vec{r})\to\psi(x,y,z)=e^{i(xk_x+yk_y+zk_z)}$$
Of course, these may not be (and in fact probably aren't) the easiest coordinate systems to use for performing these Taylor expansions, but the same idea applies - pick a coordinate system and expand as you would with a normal multidimensional function (but be careful with periodic coordinates, as they make things more complicated to interpret if you choose to Taylor-expand in that direction). The neat thing is that Taylor expansions in different coordinates will tell you somewhat different things about a function's behavior, since different coordinates become small in different regions of space. For example, you might want to expand only in the magnitudes $r$, $r_0$, $k$ of each of the vectors, in which case you would get:
$$\psi(\vec{r})\to\psi(r,\theta)=\left(\sqrt{r^2+r_0^2-2rr_0\cos\theta}\right)^{3/2}$$
$$\psi(\vec{r})\to\psi(r,\theta)=e^{irk\cos\theta}$$
where $\theta$ is, in the first example, the angle between $\vec{r}$ and $\vec{r_0}$ (in other words, the axis of rotational symmetry of the function is located along $\vec{r_0}$), and is, in the second example, the angle between $\vec{r}$ and $\vec{k}$ (so that the axis of rotational symmetry is along $\vec{k}$).
