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
$$\sum_{n = m}^{\infty} \lvert f_n(z)\rvert \leqslant \frac{1}{m} + \sum_{n = m}^{\infty} 2^{-n} = \frac{1}{m} + 2^{1-m}\,.$$
For (i), given $\epsilon > 0$ there exist $N \in \mathbb{N}$ such that for all $m > n > N$ we have for all $x \in X$
$$\left|\sum_{k=n+1}^m \log (1 + g_k(x)) \right| \leqslant\sum_{k=n+1}^m |\log (1 + g_k(x))| \leqslant \frac{3}{2}\sum_{k=n+1}^m|g_k(x)| < \epsilon,$$
since the RHS series is uniformly convergent.
For (iii), the series $\sum_{n \geqslant 1} \log(1+g_n(x))$ converges uniformly if and only if $\sum_{n \geqslant n_0+1} \log(1+g_n(x))$ converges uniformly. We can add or subtract a finite number of terms without consequence.
Also if $S_n(x) \to S(x)$ uniformly then $\exp(S_n(x)) \to \exp(S(x))$ since the exponential function is continuous everywhere. Thus, uniform convergence of $h(x)$ imples uniform convergence of $f(x)$.
Addendum: Absolute convergence of an infinite product implies convergence
Let $P_n = \prod_{k=1}^n (1+a_k)$ and $Q_n = \prod_{k=1}^n (1+|a_k|)$. We have
$$P_n - P_{n-1} = (1+a_1) \ldots (1+a_{n-1}) a_n, \\ Q_n - Q_{n-1} = (1+|a_1|) \ldots (1+|a_{n-1}|) |a_n|,$$
and it follows that $|P_n - P_{n-1}| \leqslant Q_n - Q_{n-1}$.
If $\prod(1+|a_n|)$ is convergent then the series $\sum(Q_n- Q_{n-1})$ converges since
$$\lim_{N \to \infty}\sum_{n=2}^N (Q_n - Q_{n-1})= Q_1 + \lim_{N \to \infty}Q_N = \prod_{n=1}^\infty (1 + |a_n|)$$
By the comparison test, the series $\sum(P_n- P_{n-1})$ is convergent and, therefore, the product $\prod(1+a_n)$ is convergent since
$$\prod_{n=1}^\infty(1+a_n) = \lim_{N \to \infty} P_N = P_1 + \sum_{n=2}^\infty (P_n - P_{n-1})$$
A final but important detail is to show that $\lim_{N \to \infty}P_N \neq 0$. This follows from the convergence of $\sum|a_n|$ which implies $1 + a_n \to 1$. It follows that the series $\sum |a_n(1+a_n)^{-1}|$ and, hence, the product $\prod(1 - a_n(1+a_n)^{-1})$ are convergent. Thus,
$$\lim_{N\to \infty} \frac{1}{P_N} = \prod_{n=1}^\infty \frac{1}{1+a_n} = \prod_{n=1}^\infty \left(1 - \frac{a_n}{1+a_n}\right) \neq \infty$$
Best Answer
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 $\{z : |z| \le M\}$.