Find the linear fractional transformation that maps the circles |z-1/4| = 1/4 and |z|=1 onto two concentric circles centered at w=0?

I am very close to the solution I think. Since the circles cross the real axis, I want to find mappings from $z \to w$ such that $1/2 \to m$ and $0 \to -m$ and $-1 \to n$ and $1 \to -n$. Using $$ w = \frac{az+b}{cz+d}$$ we get $$ \frac{a+2b}{c+2d} = -\frac{b}{d}$$ and $$ \frac{b-a}{d-c} = - \frac{a+b}{c+d}$$ but where do we go from here? I am a little lost in the simplifying algebra here to solve for the $a,b,c,d$ in the transformation.

Starting from your ansatz, we can reduce the degrees of freedom by positing that we want to map $1 \mapsto 1$ and $-1\mapsto -1$. If your ansatz works at all (it does, but one needs to know a bit about Möbius transformations to know that a priori), we can achieve that by afterwards dividing by $n$.

That gives us the relations

\begin{gather} a+b = c+d\\ a-b = d-c\\ (a+2b)d = - (c+2d)b \end{gather}

for the desired transformation. From the first two, we obtain $a = d$ and $b = c$, so inserting that into the third, we get

$$(a+2b)a = - (b+2a)b,$$

and that becomes

$$a^2 + 4ab + b^2 = 0,$$

or

$$(a+2b)^2 = 3b^2.$$

We can normalise $b = 1$ (since $b = 0$ would lead to $a = 0$ and that doesn't yield a Möbius transformation), then $(a+2)^2 = 3$ gives us $a = -2 \pm \sqrt{3}$.

Let's pick $a = -2-\sqrt{3}$. So we look at

$$T(z) = \frac{(-2-\sqrt{3})z + 1}{z -(2+\sqrt{3})}.$$

That gives us

$$\begin{aligned} T(1) &= \frac{-1-\sqrt{3}}{-1-\sqrt{3}} = 1; & T(-1) &= \frac{3+\sqrt{3}}{-3-\sqrt{3}} = -1;\\ T(0) &= \frac{1}{-(2+\sqrt{3})} = -(2-\sqrt{3}); & T\left(\tfrac{1}{2}\right) &= \frac{-\sqrt{3}}{1-2(2+\sqrt{3})} = \frac{-\sqrt{3}}{-\sqrt{3}(2+\sqrt{3})} = 2-\sqrt{3}; \end{aligned}$$

and the four points are mapped as desired. Since $T(\mathbb{R}\cup \{\infty\} = \mathbb{R} \cup \{\infty\}$, and the original two circles intersected the real line at right angles, the two image circles also do that, hence their centres are real, and since the intersections of the image circles with the real line are symmetric with respect to $0$, the centre of both is $0$.