= Solution
Let $\phi_i:D\to U_i$ be the normalized maps, and put $V=\phi_1(U_2)$. A simple loop in $U_1$ has its interior there, so applying $\phi_1^{-1}$ preserves the property of surrounding $0$. There is a disjoint decomposition
$$
\mathbf L_D\setminus\mathbf L_V
=(\mathbf L_D\setminus\mathbf L_{U_1})
\ \dot\cup\ (\mathbf L_{U_1}\setminus\mathbf L_V).
$$
The conformal restriction property identifies the measure of the second set with $\nu(\mathbf L_D\setminus\mathbf L_{U_2})$. Therefore
$$
\boxed{A^\nu(\phi_1\circ\phi_2)=A^\nu(\phi_1)+A^\nu(\phi_2).}
$$
The composition fixes $0$ and has positive derivative there, so it is exactly the normalized map onto $V$. No interchange of infinite differences is needed: the argument is additivity on disjoint sets. The later logarithmic expression is consistent with this identity because $(\phi_1\circ\phi_2)'(0)=\phi_1'(0)\phi_2'(0)$.
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