We know that the exponential function gives a biholomorphism between an open strip parallel to the real axis with a width $\leqslant 2\pi$ and an angular sector. One angular sector that is particularly easy to handle, especially considering we want to reach the unit disk, is the upper half plane. We need a width of $\pi$ for that, so

$$f_1(z) = e^{\pi z}$$

gives a conformal mapping of $U$ to the upper half plane, and it maps the $y = 0$ part of the boundary to the positive half-axis, and the $y = 1$ part to the negative half-axis.

That looks promising, now we need a biholomorphic map between the upper half-plane and the unit disk that maps the positive half-axis to the part of the unit circle in the lower half-plane, and the negative half-axis to the part in the upper half-plane. That means it must map $\{0,\infty\}$, the remaining part of the boundary of the upper half-plane, to $\{-1,1\}$, the remaining part of the boundary of the unit disk. With the boundaries properly oriented, both regions lie to the left of their respective boundaries, so when the positive half-axis is traversed from $0$ to $\infty$, the lower unit semicircle is traversed from $-1$ to $1$, hence the biholomorphism must map $0 \mapsto -1$ and $\infty \mapsto 1$. The Möbius transformation

$$f_2(z) = \frac{z-i}{z+i}$$

is easily seen to be the solution to that problem.

We need to compose with $f_1^{-1} \circ f_2^{-1}$, so let's compute the inverses and compose:

$$f(z) = \frac{1}{\pi} \log \left(i\frac{1+z}{1-z}\right)$$

biholomorphically maps the unit disk to the strip $0 < \operatorname{Im} z < 1$, mapping the unit semicircle in the lower half plane to the real axis, and the upper unit semicircle to the line $\operatorname{Im} z = 1$.