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,
$$\lim_{N\to \infty} \frac{1}{P_N} = \prod_{n=1}^\infty \frac{1}{1+a_n} = \prod_{n=1}^\infty \left(1 - \frac{a_n}{1+a_n}\right) \neq \infty$$
Yes, $\sum_{n=1}^\infty f_n= f$ is a sum of functions and is defined by saying that, for a specific value of x, f(x) is the value of the numeric sum $\sum_{n=1}^\infty f_n(x)$.
"And we specify whether we mean uniformly or pointwise". Not quite! Convergence of a sequence of series of functions is always "pointwise". It may then "converge uniformly" or not. The definition of convergence of "$\lim_{n\to\infty} f_n= f$" is that the numerical sequence "$\lim_{n\to\infty} f_n(x)$" converges to "$f(x)$" for every x in the domains of all the $f_n$. That is "pointwise" convergence- it converges at every "point"- every x-value.
We can then determine whether or not that convergence is also uniform.
Determining "pointwise" convergence is exactly the same as for numerical sequence. A sequence of functions, $f_n$, converges pointwise to f if and only if the numerical sequence, $f_n(x)$, converges to f(x) for every x in the domain. That is, if for a specific x, given $\epsilon> 0$ there exist N such that if n> N then $|f_n(x)- f(x)|< \epsilon$. This is done at every value of x- given an $\epsilon$, what N will work may be different for different x values.
The sequence of functions converges uniformly (on some interval) if, in addition to converging pointwise, given $\epsilon$ there exist an N that will work for all x in that interval.
Notice I was talking about sequences here. A series, $\sum_{n= 1}^\infty f$ converges to f if and only if the sequence of finite sums, $\sum{n=1}^N f_n$ converges to f so the same distinction between "pointwise" and "uniform" convergence applies. A series, if it converges, always converges "pointwise". It them may or may not converge "uniformly".
Best Answer
In finite dimensions, all norms on a vector space are equivalent. Thus we tend to stick with the well-understood Euclidean norm on $\Bbb R^n$, occasionally pulling out the sup or taxicab norms when computationally convenient.
But most function spaces are infinite-dimensional, which means norms finally have enough freedom to disagree with each other. And the thing is, there are several different norms that are useful. And when as here, we are talking about several possible spaces, each with its own norm, it becomes confusing which norm is being referred to.
There are only two notations for norms that are so widely used that an author can use them without explaining. These are:
In specialized fields of mathematics, there are some other notations that are well-known in that field, and may be used when speaking to those well-versed in the field. But when speaking to a wider mathematically inclined audience, the notations above are the only ones that can safely be used without explanation. And of course even then, there will always be a few novices who might need an introduction.
The tip-off that Daniel Fischer is correct about the intent here is the definition specifies that $\sum_{n=1}^\infty \|f_n\|$ must converge uniformly. If $\|f_n\|$ meant the supremum norm of $f_n$, then the series would be a sum of real numbers, and talking about "uniform" convergence makes no sense. Uniform with respect to what? In order to talk about uniform convergence, $\sum \|f_n\|$ must be a sum of functions. The interpretation that makes $\|f_n\|$ a function is that it means the function $$\|f_n\| : S \to \Bbb R : x \mapsto \|f_n(x)\|$$