Your composition unfortunately does not work out. We have
$$\Omega_1 = f_1(\Omega) = \{z : \operatorname{Im} z > -1\} \setminus \overline{\mathbb{D}},$$
and thus
$$\Omega_2 = f_2(\Omega_1) = \{z : \operatorname{Re} z > -1\} \setminus \overline{\mathbb{D}},$$
and both $f_1$ and $f_2$ are biholomorphic. But then $f_3$ is not injective on $\Omega_2$. While indeed $f_3(\Omega_2) = \mathbb{C}\setminus \overline{\mathbb{D}}$, the mapping is only locally conformal, not conformal. And $\mathbb{C}\setminus \overline{\mathbb{D}}$ is not simply connected, hence there is no conformal map from that onto the upper half-plane (you get a conformal map $\widehat{\mathbb{C}}\setminus \overline{\mathbb{D}} \to \mathbb{H}$ from $f_4$, though).
To get a conformal map $\Omega \to \mathbb{H}$, the usual way is to start by using a Möbius transformation to map the point of contact of the two boundary curves to $\infty$. That maps the two boundary curves to parallel lines (since both are circles or straight lines), and $\Omega$ to a parallel strip. Mapping a parallel strip conformally to a half-plane is a standard exercise.
Taking $g_1(z) = \frac{1}{z}$ for the Möbius transformation, the real line is mapped to itself (interchanging $0$ and $\infty$), and the circle $\left\lvert z - \frac{i}{2}\right\rvert = \frac{1}{2}$ is mapped to the line $\operatorname{Im} z = -1$, and $\Omega$ to the parallel strip $\{ z : -1 < \operatorname{Im} z < 0\}$ between these two lines.
The map $g_2(z) = \exp\left(\pi\left(z + \frac{i}{2}\right)\right)$ conformally maps the strip to the right half-plane, and a rotation then maps the right half-plane to the upper. Altogether, we obtain the conformal map
$$F \colon z \mapsto i\cdot \exp\left(\pi\left(\frac{1}{z}+\frac{i}{2}\right)\right)$$
from $\Omega$ to $\mathbb{H}$.
A Möbius transformation that maps the upper half plane to itself must map the boundary, i.e. the real line, to itself as well. With the standard definition for the transformation,
$$f(z) = \frac{az+b}{cz+d},$$
it is fairly straightforward to see that the parameters $a, b, c, d$ must be real to map the real line to itself (try plugging in $z=0$, $z=\infty$, and then one other real number).
Then since $f$ is supposed to interchange $z_1$ and $z_2$, that means applying $f$ twice should send $z_1$ back to itself, and the same for $z_2$. Note that $f(f(z))$ is again another Möbius transformation with real coefficients. You can now solve the equation for the fixed points of the transformation (or look up the formula). This should answer the question about whether the transformation is possible.
Best Answer
The points in the unit circle satisfy $|z|<1$. Thus when $z_0$ is in the upper half plane, points in the upper half plane will be closer to $z_0$ than to $\bar{z_0}$, that is $|z-z_0|<|z-\bar{z_0}|$. This means that the fraction in $T(z)$ is less than $1$ precisely when $z$ is in the upper half plane. (The $e^{i\theta}$ factor has magnitude one and thus is only a rotation.)