One goes about understanding a topology by looking at the basis, the "simplest" open sets of which all other sets are composed as unions. For a metric space, the basis consists of the balls, of course. And what's even nicer about topological vector spaces is that you only have to study what the basis sets look like at the origin, since topological vector spaces are shift-invariant.
The norm-topology's basis at the origin consists of the balls $ B(0,r) $. Easy enough. What do basis sets look like in the weak topology? Take a bounded linear functional $ f \in X^* $, and then look at $ \{ x \in X : |f(x)| < r \} $. You can hopefully imagine, this is "the space that is sandwiched between two hyperplanes". Now, this isn't quite the basis. To get a basis (at the origin), you need to consider all possible intersections of these "sandwiches". But the important thing is, you only get to "sandwich" in a finite number of directions $ f_1, \ldots, f_N $. So, in the infinite dimensional case, there's always some direction that you fail to sandwich. Hence, these basis sets are not bounded. This is counterintuitive, because in $ \mathbb{R}^3 $ (for example), you obtain a bounded set once you "sandwich" along the x-axis, y-axis, and z-axis (or in 3 directions of your choice).
Question 1 Let's first get clear on why the norm-topology contains the weak topology. Let $ U $ be open in the weak topology. Fix a point $ x_0 \in U $. We can find a basis set $ V $ with $ x \in V \subseteq U $. Recall, by "basis set", I mean a finite number of sandwiches, but we can be rigorous about this: for some numbers $ m_1 < M_1, \ldots m_N < M_N $ and $ f_1, \ldots f_N \in X^* $, the basis set $ V $ can be expressed
$$
V = \{ x \in X : m_1 < f_1(x) < M_1, \ldots, m_N < f_N(x) < M_N \}
$$
Now, it's pretty straightforward to find a ball $ B(x_0, r_0) $ that fits inside of $ V $. Do you see how? It relies on the boundedness of hte functionals $ f_1, \ldots f_N $. Thus we have
$$
x_0 \in B(x_0, r_0) \subseteq V \subseteq U
$$
which, you'll recall, shows that $ U $ is open in the norm topology (make sure you see why).
Question 1, Part b You then asked why $ x \mapsto |f(x)| $ is norm-continuous. But this is a composition of two continuous functions, the absolute value function and $ f \in X^* $ (which is continuous by definition).
Question 2 You asked about showing that $ g : x \mapsto \|Lx\| $ is weakly continuous, where $ L:X \rightarrow X $ is a bounded linear map. Surprisingly, it is not continuous with respect to the weak topology (in general). Indeed, pick $ X $ to be your favorite infinite-dimensional space, let's say $ X = \mathcal{l}^1(\mathbb{N}) $ and $ L = Identity : X \rightarrow X $. Then I claim $ g : x \mapsto \|Lx\| $ is not weakly continuous, since indeed $ g^{-1}((-a,a)) = B(0,a) $ is bounded, whereas weakly open sets in $ \mathcal{l}^1 $ are not bounded (!).
Addendum Fundamentally, the thing that's surprising here is that the norm $ x \mapsto \|x\| $ is not weakly continuous. To try to wrap our heads around this, consider the weak basis we discussed above, composed of finite intersections of "hyperplane sandwiches." This basis of the weak topology tells us that, in order for a function $ f : X \rightarrow \mathbb{R} $ to be weakly continuous, it is only allowed to change in a finite number of directions. The norm function, on the other hand, changes in all directions that you walk away from the origin.
It seems that what the authors of the paper are saying is that the topology if total variation in the space of finite measures is to restrictive. The weak topology they are referring there is the weak topology $\sigma(\mathcal{M},\mathcal{C}_b(X))$ where $\mathcal{M}$ is the space of finite measures, and $\mathcal{C}_b(X)$ is the space of continuous bounded functions (on $X$). A local base of the topology $\sigma(\mathcal{M}(X),\mathcal{C}_b(X))$ is given by sets of the form
$$V(f_1,\ldots,f_n;\varepsilon)=\{\mu\in\mathcal{M}:|\mu(f_j)|<\varepsilon\}$$
where $\mu(f_j)=\int_Xf_j\,d\mu$, $f_1,\ldots,f_n\in\mathcal{C}_b(X)$, $n\in\mathbb{N}$.
In this topology, a net $(\mu_\alpha:\alpha \in D)$ converges to $\mu$ iff for any $f\in\mathcal{C}_b(X)$
$$\lim_\alpha \mu_\alpha(f)=\mu(a)$$
I assume for the rest of this posting that $X$ is a complete separable metric space.
As for your questions:
If $\mu_n$ converges to $\mu$ in total variation, then $\mu_n$ converges to $\mu$ in $\sigma(\mathcal{M}(X),\mathcal{C}_b(X))$. Indeed, for $f\in\mathcal{C}_b(X)$
$$\Big|\int_Xf\,d\mu_n-\int_Xf\,d\mu\Big|\leq\int_X|f|\,d|\mu_n-\mu|\leq\|f\|_u\|\mu_n-\mu\|_{TV}\xrightarrow{n\rightarrow\infty}0$$
Consider a sequence $x_n\in X\setminus\{x\}$ such that $x_n\xrightarrow x$ in $X$. Consider the measures $\delta_{x_n}$ and $\delta_x$ where $\delta_{x_n}(A)=\mathbb{1}_{A}(x_n)$ (similar for $\delta_x$). You can easily check that $\mu_n$ converses weakly to $\mu$ (i.e. in the $\sigma(\mathcal{M}(X),\mathcal{C}_b(X))$ topology); however,
$\|\delta_{x_n} -\delta_x\|_{TV}=1$
I leave this for the OP to think about.
Best Answer
1: It is merely notational.
2: The sets you’ve written are neighbourhoods, but you’ve written a subbasis, not a basis. What I mean by that is that you need to allow non-fixed $x$ to get the “finite intersection of open sets is open” criterion.
Likewise we need to change the expression for the weak basis:
3: Linear functionals have a kernel with codimension $1$. If I want to squeeze $S,T$ into the same weak neighbourhood where particular $\psi,x$ are fixed, I need to make sure that $S-T$ has small image along the support of $\psi$ - they may create a huge vector along the kernel. The same goes for multiple $\psi$: we intersect their kernels, we will still (when $Y$ is infinite dimensional) have an enormous vector subspace on which they are all $0$. I need only require $S-T$ is small outside of this subspace, but within the shared kernel? It can be huge. This is similar to the fact that every weak open neighbourhood in the weak topology on $X$ is unbounded when $X$ is infinite dimensional.