I have some lecture notes containing the following.
An analytic function $F$ in the disc is outer if there is an $h\ge0$ s.t $\int_{\partial \mathbb{D}} \mid log h \mid < \infty$ and $F(z)=e^{\int_{\partial \mathbb{D}} \frac{x+z}{x-z}log h(x)dx}$
In this case $F$ is the outer function whose modulus equal $h$ a.e on $\partial \mathbb{D}$.
According to me this suggests that $F$ in the second part of the theorem is of the form
$F(z)=e^{\int_{\partial \mathbb{D}} \frac{x+z}{x-z}log(log \mid f \mid) dx}$.
This does not lead me to the conclusion of the second part of the theorem. If we have $F(z)=e^{\int_{\partial \mathbb{D}} \frac{x+z}{x-z}log \mid f \mid dx}$ then atlest I get $f/F=1$ a.e on $\partial \mathbb{D}$ which is mentioned as consequence right after this proof. And I feel somewhat closer to prove the $F_n$ part, with $\mid F_n \mid =\mid f \mid + \frac{1}{n}$
Should it be $\mid F \mid = \mid f \mid$ a.e on $\partial \mathbb{D}$ in the hypoteses of the second part?

Here are two classical references for this theorem:
There is indeed a typo in the second part of your quoted theorem: instead of saying "if $F$ is the outer function with $F|=\log|f|$ m-a.e. on $\mathbb{T}$" it should say something like "if $F$ is the outer function of $f$" if you follow the terminology of Rudin. Of course, as you point out, this implies that $|F|=|f|$ m-a.e. on $\mathbb{T}$.