It's not sequential because its closed subspace $M[0,1] = (C[0,1])^\ast$ is not sequential.
Here is an example of a set $A \subset M[0,1]$ that is sequentially $\tau_v$-closed but not $\tau_v$-closed:
Consider sequence of functions $f_n \in C[0,1]$, $\Vert f \Vert_{C[0,1]} = 1$ and $\operatorname{span} \{f_n\}$ is norm-dense in $C[0,1]$, and take
$$A := \bigcup_{n \ge 1} \left\{ \mu : \intop f_n d \mu = n \right\},$$
On the one hand, it's sequentially $\tau_v$-closed. Indeed, any $\tau_v$-convergent sequence is bounded in the total variation norm by the uniform boundedness principle, and $A$ intersects any ball at only finitely many closed hyperplanes. On the other hand, $0 \notin A$ but $0$ is contained in the $\tau_v$-closure of $A$ because any basic $\tau_v$-neighborhood of $0$ of the form $\left\{ \left| \intop g_i d \mu \right| < \varepsilon, i = 1, \dots, n \right\}$ intersects $\left\{ \intop f_m d \mu = m \right\}$ for any $f_m \notin \operatorname{span} \{ g_i\}$.
It's Lusin.
One can "encode" a measure on $\mathbb{R}$ by a sequence of compatible finite measures on $[-n, n]$. Now, on finite measures on $[-n, n]$ the weak topology can be strenghthened to a Polish space topology by adding total variation norm balls around $0$ as additional open sets. Here we are using the fact that on balls the weak topology becomes metrizable. For example, one can use the following metric:
$$\rho(\mu, \nu) := \sum_m \frac{1}{2^m} \arctan \left| \intop f_m d (\mu - \nu) \right| + \left| \Vert \mu \Vert - \Vert \nu \Vert \right|$$
where $\Vert \cdot \Vert$ is the total variation norm and $f_m$ is a dense sequence of functions of norm $1$.
The condition that the measures on $[-n, n]$ are compatible introduces a closed subset in the countable product of spaces of measures on $[-n, n]$. So this way we can cook up a Polish space topology on $M(\mathbb{R})$.
The argument you give for the equality of the weak operator and weak-$^*$ (or $\sigma$-weak) topologies also shows that the strong and $\sigma$-strong topologies are equal, and similarly for the strong-$^*$ and $\sigma$ strong-$^*$. This equality of topologies holds for all von Neumann algebras in standard form.
It's not hard to see that the adjoint on $L^\infty(X)$ is continuous with respect to the strong operator topology and thus the strong operator topology and strong-$^*$ topologies are equal as well. In general, this is equivalent to the finiteness of a von Neumann algebra.
Therefore, you are only left with a few distinct topologies: the weak operator topology, the strong operator topology, and the Mackey topology. It's easiest to describe the behavior of these topologies on the unit ball, and it's a result of Akemann that the Mackey topology is equal to the strong-$^*$ topology on the unit ball, or in our case the strong topology, which brings us down to two topologies: the weak and strong operator topologies.
To most easily describe the behavior of these topologies on bounded sets, it's best to replace them with the finest locally convex topologies that agree with them on bounded sets, the bounded weak-$^*$ (or equicontinuous weak-$^*$) topology and Mackey topology. This identification follows from the Mackey-Arens and Banach-Dieudonné theorems from locally convex space theory. In this case, these topologies are the topologies of uniform convergence on norm compact and weakly compact absolutely convex subsets of $L^1(X)$ respectively.
The norm compact absolutely convex subsets of a Banach space are just the absolute convex hulls of null sequences, so the bounded weak-$^*$ topology is equal to the topology of uniform convergence on null sequences in $L^1(X)$. I don't know if this counts as an explicit description. Maybe it's possible to work out something better.
If $X$ is finite, the Dunford-Pettis Theorem identifies the relatively weakly compact subsets of $L^1(X)$ with the bounded equiintegrable sets. Thus, if $X$ is finite then the Mackey topology and strong topology are equal on bounded sets to the topology of convergence in measure. This contradicts the claim in your link, which states it for the $\sigma$-weak topology. For more general localizable measure spaces, you have to consider the topology of convergence in measure on subsets of $X$ with finite measure. The proof is similar to pretty much any other proof that represents an $L^\infty$ space as a projective limit of $L^\infty$ spaces of finite measure spaces.
The weak and strong operator topologies only agree (even on the unit ball) when $X$ is purely atomic, as this is precisely when every weakly compact subset of $L^1(X)$ is norm compact.
I've never seen a good description of these topologies on unbounded subsets of $L^\infty(X)$, and it seems that it would be difficult to give one. For example, if $X$ is a separable measure space then $L^\infty(X)$ has a separable predual, and thus all of the topologies mentioned here are metrizable on bounded sets, but they are never metrizable when $X$ is infinite. This precludes the Mackey topology from actually being equal to the topology of convergence in measure, because the latter is metrizable.
An interesting book on some of these topologies is Saks Spaces and Applications to Functional Analysis by J.B. Cooper. Since this was migrated from MSE, please let me know if you want more precise references for some of the things I've mentioned.
Best Answer
First, I will describe three topologies on $C^\infty(M,N)$ for smooth manifolds $M$ and $N$, and then relate them to the topologies 1, 2, and 4 from the question. For $N=\mathbb{R}$, all of the below statements relating the topologies remain valid if we pass to the subspaces of compactly supported functions.
It seems that terminology is not completely standardized, I will call these three topologies the weak $C^\infty$-topology, the Whitney $C^\infty$-topology, and the strong $C^\infty$-topology. Assume we are given
all with the same indexing set. Consider the set of smooth $g:M\to N$ such that for all indices $i$, we have $g(K_i)\subseteq V_i$ and
$$\|D^\alpha(\psi_i g\phi^{-1}_i - \psi_i h\phi^{-1}_i)\|(m) < \varepsilon_i$$
for all $m\in K_i$ and multiindices $\alpha$ of order at most $k_i$. Sets of this form are a basis for the strong $C^\infty$-topology on $C^\infty(M,N)$. If we additionally require that the $k_i$ are bounded, then we get the Whitney $C^\infty$-topology, that is, we would not have needed a family of $k_i$, but just one non-negative integer $k$. If we require that the indexing set is finite, we get the weak $C^\infty$-topology.
If $M$ is compact, all three topologies agree, and if $M$ is not compact, they all differ. The strong topology is stronger (finer) than the Whitney topology, and the Whitney topology is stronger than the weak topology. Note that for $l< \infty$, the $C^l$-versions of the strong topology and the Whitney topology are equal, which may be why the words "strong" and "Whitney" are sometimes used interchangably. I took this way of differentiating between them from Chris Schommer-Pries's lecture notes, which contains a lot of information about these and other topologies. These two topologies are much more similar to each other than to the weak topology. For example, the Whitney and strong $C^\infty$-topologies have the same weak homotopy type, because they even have the same convergent sequences.
Using these definitions, I do not know whether you meant the Whitney or the strong $C^\infty$-topology in 2. In any case, your first guess is correct, as can be seen from the above description. For $N=\mathbb{R}$, the Whitney $C^\infty$-topology is equal to topology 4. To see this, express the vector fields as linear combinations of partial derivatives in coordinates. Also, it is possible to find a smooth $\varepsilon$ which takes values at most $\varepsilon_i$ on $K_i$ for all $i$ by using bump functions and that $U$ was assumed to be locally finite.
In addition to the above link, see Hirsch's "Differential Topology" and Golubitsky's and Guillemin's "Stable Mappings and Their Singularities" for more information about the weak and the Whitney $C^\infty$-topologies, which use the more intrinsic language of jet bundles in addition to something like the above.