This question regards using a Taylor polynomial to approximate the expression
$$V_P = \frac{kq}{\sqrt{r^2-2ra\cos{\theta}+a^2}}\tag{1}$$
We can rewrite the right-hand side as
$$\frac{kq}{\sqrt{r^2(1-\frac{2a\cos{\theta}}{r}+\frac{a^2}{r^2})}}$$
$$=\frac{kq}{r\sqrt{1+\epsilon_1}}\tag{2}$$
where
$$\epsilon_1=-\frac{2a\cos{\theta}}{r}+\frac{a^2}{r^2}\tag{3}$$
Assume we want to investigate what happens when $r>>a$. Then $\frac{a}{r}\to 0$, and $\epsilon_1\to 0$.
We can approximate $\frac{1}{\sqrt{1+\epsilon_1}}$ using a first-order Taylor polynomial at $0$
$$\frac{1}{\sqrt{1+\epsilon_1}}\approx 1-\frac{\epsilon_1}{2}\tag{4}$$
so that
$$V_P \approx \frac{kq}{r}\cdot (1-\frac{\epsilon_1}{2})\tag{5}$$
$$=\frac{kq}{r}\cdot \left [1- \left (-\frac{2a\cos{\theta}}{r}+\frac{a^2}{r^2}\right ) \right ]\tag{6}$$
$$=\frac{kq}{r}\cdot \left (1+\frac{2a\cos{\theta}}{r}-\frac{a^2}{r^2} \right )\tag{7}$$
An Alternative Approximation
Okay, but what happens if instead of approximating $\frac{1}{\sqrt{1+\epsilon_1}}$ we approximate just $\sqrt{1+\epsilon_1}$?
We have that if $f(x)=\sqrt{1+x}$ then near $x=0$ we have
$$f(x)\approx 1+\frac{x}{2}\tag{8}$$
Can we also say then that $$\frac{1}{\sqrt{1+x}}\approx \frac{1}{1+\frac{x}{2}}=\frac{2}{2+x}\tag{9}$$
If so, then we obtain
$$V_P\approx \frac{kq}{r}\cdot\frac{2}{2+\epsilon_1}\tag{10}$$
$$=\frac{2kqr}{-2ar\cos{\theta}+a^2+2r^2}\tag{11}$$
Questions
Are the steps in this alternative approximation correct (in particular, is (9) correct)?
If so, are (7) and (11) both correct approximations for $V_P$?
I don't see any mistakes in your two approximations. To give a more general answer to your question:
Let us say you have a function $f :A \longrightarrow B$, as well as two approximations $f_1,f_2 : A\longrightarrow B$, so that all of them live in the same function space $(X,d)$ with metric $d$ and also $d(f,f_1),d(f,f_2)\leq \varepsilon$ for some $\varepsilon>0$. Then, due to the triangle inequality, one also has $d(f_1,f_2) \leq d(f_1,f)+ d(f_2,f) \leq 2\varepsilon$.
Thus, if both approximations are "close" to $f$ (with respect to the chosen metric $d$), they can't be very different (again, with respect to $d$). However, there is no reason why they should be the same.