Take any $\overline{B(z_0,r)} \subset D$, then $f_n$ is uniformly bounded on $\overline{B(z_0,r)}$. Hence $f_n'$ is uniformly bounded, say $|f_n'| \le M, z \in \overline{B(z_0,r)}$.
For any $\epsilon >0$, take $\delta=\frac{\epsilon}{3M},$ for $|z_1-z_2|<\delta$ implies that $$|f(z_1)-f(z_2)| =|\int_{z_2}^{z_1}f_n'(z)dz|< \frac{\epsilon}{3}$$
Hence $f_n$ equicontinuous on $B(z_0,r)$.
Take any compace subset $K \subset \Omega$, we have a finite cover $$K \subset \bigcup_{j=1}^l B(z_j,\frac{\delta}{2})$$
So $f_n$ also equicontinuous on $K$.
Since $$\lim_{n \to \infty}f_n(z_j)$$ exists, then for any $\epsilon >0$, there exists a $N_j \in \mathbb{Z_+}$, for $n,m>N_j$ implies that $|f_n(z_j)-f_m(z_j)| < \frac{\epsilon}{3}$.
Pick $N=max(N_1,……,N_j)$, for $n,m>N$, we have $$|f_n(z)-f_m(z)|<|f_n(z)-f_n(z_k)|+|f_n(z_k)-f_m(z_k)|+|f_m(z_k)-f_m(z)|<\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}=\epsilon$$
Here $z \in B(z_k, \frac{\delta}{2})(1 \le k \le l)$, and so $f_n$ uniformly converges on any compact subset of $\Omega$.
Your series is
$$
\sum_{n=1}^\infty\left(\big|w^{2n-1}e^z\big|+\big|w^{2n-1}e^{-z}\big|\right)
$$
For $|w|<1$ a fixed number and $z$ in a bounded set, this should be easy to bound.
Suppose $z$ is in the set $\{z : |z| \le M\}$. Then
$|e^z| = e^{\operatorname{Re} z} \le e^M$ and
$|e^{-z}| = e^{-\operatorname{Re} z} \le e^M$, so
$$
\sum_{n=1}^\infty\left(\big|w^{2n-1}e^z\big|+\big|w^{2n-1}e^{-z}\big|\right)
\le
\sum_{n=1}^\infty\left(|w|^{2n-1}e^M+|w|^{2n-1}e^{M}\right)
= \frac{2e^M |w|}{1-|w|^2} .
$$
Therefore, by the Weierstrass M-test, the infinite series converges uniformly on the set $\{z : |z|\le M\}$. And so $\theta(z)$ converges uniformly and absolutely on $\{z : |z| \le M\}$.
Best Answer
We can also construct counterexamples using holomorphic functions. Let $K$ be the closed unit disk, define $\zeta_n = \exp(i/n)$ for $n \in \mathbb{N}\setminus \{0\}$, and $h(z) = \frac{1}{2}(1 + z)$. Choose positive integers $m_n$ such that $$\lvert h(z)\rvert^{m_n} \leqslant 2^{-n}$$ for all $z \in K$ with $\lvert z-1\rvert \geqslant \frac{1}{10n^2}$ and put $$f_n(z) = \frac{1}{n}h(\zeta_n z)^{m_n}\,.$$ Then $\lVert f_n\rVert = \frac{1}{n}$, so the series is not normally convergent on $K$. But it is absolutely and uniformly convergent on $K$. For every $z \in K$, we have $\lvert f_n(z)\rvert \leqslant 2^{-n}$ for all but at most one $n$, so $$\sum_{n = m}^{\infty} \lvert f_n(z)\rvert \leqslant \frac{1}{m} + \sum_{n = m}^{\infty} 2^{-n} = \frac{1}{m} + 2^{1-m}\,.$$