It seems hard to me to give an "answer" to this quesiton, but let me try to share some thoughts. On a Riemannian manifold, the Riemannian metric can be used to define a distance function (sometimes also referred to as a metric) which in turn defines a topology on the manifold. As you note, it is a nice feature, that this topology coincides with the manifold topology. But this also implies that the topology is completely independent of the Riemannian metric in question. For example, you can put a Riemannian metric on $\mathbb R^4$ which does not admit a single isometry. But still, this will induce the usual topology (which of course is locally homogeneous).
In the pseudo-Riemannian case, this goes wrong in several respects. If you try to define a notion of "distance" parallel to the Riemannian case, you can get positive and negative distances and (even worse) different points may have distance zero (if they lie on a lightlike geodesic). This certainly does not fit into the setting of metric spaces, and while one could try to use it to define a topology, this topology would certainly be very badly behaved. (Any point which can be reached from $x$ by a lightlike geodesic would be considered as being arbitrarily close to $x$.)
The basic point in my opinion is that in mathematics you usually look at the same object from different perspectives. This is usually expressed by looking at different classes of "(iso-)morphisms". Any Riemannian or pseudo-Riemannain manifold has an underlying smooth manifold, which in turns has an underlying topological space (and if you want to push things further you can look at the underlying measure spaces or even the underlying set). The difference between these pictures is whether you look at isometries or at diffeomorphisms (respectively smooth maps), homeomorphisms (respectively continous maps) or measurable maps and isomorphisms. Now any smooth manifold is homogeneous under its group of diffeomorphisms (and much more than that is true), whereas (pseudo-)Riemannian manifolds which are homogeneous under their isometry group are rather rare. This of course implies that the topologies of smooth manifolds are always homogeneous. Even worse, any two compact manifolds (regardless of their dimension) are isomorphic as measure spaces.
So the "answer" from my point of view would be that Minkowski space is the smooth manifold $\mathbb R^4$ endowed with a flat Lorentzian metric. There are many properties of this metric, which are by no means reflected in the topology of $\mathbb R^4$, but this is also true for Riemannian metrics on $\mathbb R^4$. There certainly are some structures resembling topologies, which can be used to encode interesting properties of Minkowski space (I don't know the Zeeman and Hawking topologies you refer to, but I would guess that they are such structures, and causal structures are another example). But I don't think that they should be used as a replacement for the vector space topology on $\mathbb R^4$ ...
Let me finally remark that Minkowski space is homogeneous as a pseudo-Riemannian manifold (since translations are isometries). They are not homogeneous on an infinitesimal level, since there are different directions emanating from a point. (This is another bit of confusion. Minkowski space should not be considered as a vector space endowed with an inner product. Otherwise there would be a distinguished point - the origin. Mathematically speaking it is an affine space togehter with the inner product on each tangent space.)
Initially, Stiefel-Whitney classes are only defined for smooth manifolds (if the manifold is not smooth, it doesn't have a tangent bundle). However, Wu's theorem states that $w = \operatorname{Sq}(\nu)$ and we can use this as a definition of Stiefel-Whitney classes for topological manifolds; in particular, $w_4(M)$ is defined even if $M$ does not admit a smooth structure or even a PL structure.
Freedman's Theorem states every non-degenerate bilinear form $b$ arises as the intersection of a closed, simply connected four-manifold. Moreover, if $b$ is even, there is a unique such manifold up to homeomorphism, and if $b$ is odd, there are two up to homeomorphism and they are distinguished by the Kirby-Siebenmann invariant.
As $E_8$ is an even form, there is a unique (up to homeomorphism) closed, simply connected four-manifold with intersection form $E_8$; we denote it by $M_{E_8}$.
As $(1)$ is an odd form, there are (up to homeomorphism) two closed, simply connected four-manifolds with intersection form $(1)$. One is $\mathbb{CP}^2$ which has zero Kirby-Siebenmann invariant (it admits a smooth structure and hence a PL structure). We denote the other manifold by $*\mathbb{CP}^2$ and note that it has non-trivial Kirby-Siebenmann invariant.
Recall that for a closed smooth four-manifold $M$, the Stiefel-Whitney number $\langle w_4(M), [M]\rangle$ is the mod $2$ reduction of the Euler characteristic $\chi(M)$; this is Corollary $11.12$ of Milnor and Stasheff's Characteristic Classes. The same is true for topological manifolds (using the definition of Stiefel-Whitney classes outlined above), see this question.
With all of this in mind, we can now see that $\operatorname{ks}(M)$ and $w_4(M)$ are unrelated as the following table demonstrates.
$$
\begin{array}{c|cc}
M & \operatorname{ks}(M) & w_4(M)\\
\hline
S^4 & 0 & 0 \\
\mathbb{CP}^2 & 0 & \neq 0 \\
M_{E_8} & \neq 0 & 0 \\
*\mathbb{CP}^2 & \neq 0 & \neq 0
\end{array}
$$
If the four-manifold $M$ admits a PL structure, then $\operatorname{ks}(M) = 0$, but the converse is not true. For example, $M_{E_8}\# M_{E_8}$ has trivial Kirby-Siebenmann invariant but it does not admit a PL structure (every PL manifold of dimension less than or equal to $7$ is smoothable, but $M$ is not smoothable by Donaldson's theorem). The class $w_4(M)$ has nothing to do with the obstructions to admitting a PL structure (both $S^4$ and $\mathbb{CP}^2$ admit PL structures).
Best Answer
Here’s a theorem from O’Neill’s Semi-Riemannian Geometry, With Applications to Relativity:
The proof is relatively short once you accept the equivalence of (3) and (4) by referring to some other source. So, the question of smooth manifolds admitting (time-orientable) Lorentzian metrics is essentially a question about the underlying topology.
Next, suppose $(M,g)$ is a Lorentzian manifold, and let $M_{\tau}$ be the set of all time cones in all the tangent spaces; then there is a canonical projection map $\pi:M_{\tau}\to M$, and it is possible to provide $M_{\tau}$ a topology and smooth structure such that $\pi$ becomes a smooth covering map, which is 2-1 (because there are exactly two time cones for each point). Then, $\pi$ is in particular a local diffeomorphism, so the pullback $g_{\tau}:=\pi^*g$ is a Lorentzian metric and with this, $\pi: (M_{\tau},g_{\tau})\to (M,g)$ becomes a Lorentzian covering called the time-orientation covering of $(M,g)$. Now, we have the relatively easy theorem:
Btw, you can also define a smooth action of $\Bbb{Z}_2$ (doesn’t matter if it’s right or left, since it’s Abelian) on $M_{\tau}$ where $-1$ acts by sending a timecone in $T_pM$ to the opposite timecone in $T_pM$. The orbit of a given point in $M_{\tau}$ (i.e of a given timecone) is simply the set of two timecones in a given tangent space, but this is precisely the fiber of $\pi$. Hence, by the universal property of quotient manifolds, we have $M\cong M_{\tau}/\Bbb{Z}_2$, and since the $\Bbb{Z}_2$ action on $M_{\tau}$ is free (obviously), it follows $(M_{\tau},\pi,M,\Bbb{Z}_2)$ is a principal bundle. Since $M_{\tau}$ is a principal bundle, the second statement is thus equivalent to $M_{\tau}$ admitting a global section. So, the theorem is saying that existence of a smooth timelike vector field in $(M,g)$ is equivalent to a smooth choice of timecones.
Also, there’s a relatively standard topological lemma (which you can google or try to prove yourself) that for a covering (say of manifolds) $\pi:X\to B$, if the base $B$ is simply connected, then $(X,\pi,B)$ is trivializable. Thus, we have the following corollary of (2) above:
Thus, for a given Lorentzian manifold, we have a sufficient condition on the topology which ensures time-orientability. But, I’m not aware of necessary and sufficient conditions (not that I’ve looked too hard).
Finally, it should also be noted that not every Lorentzian manifold is time-orientable, and indeed, once we consider manifolds which are not simply-connected, there are examples of Lorentzian metrics on the same manifold which are time-orientable, and also ones which are not time-orientable. In fact, even if we assume that the manifold is orientable (a topological assumption as you say), that still doesn’t really help us with time-orientability. So, time-orientability of a given Lorentzian metric is a combination of both topological and metric properties.
First let’s record the following easy lemmas (which are of inherent independent interest)
By induction, it suffices to prove the case of two factors; call the manifolds $X,Y$, and let $m:=\dim Y$. If $X,Y$ are orientable, then letting $\mu_X,\mu_Y$ be respective volume forms, we have that $\mu:=\pi_X^*\mu_X\wedge\pi_Y^*\mu_Y$ will be a volume form. Conversely, suppose $X\times Y$ is orientable. Letting $Y_0$ be an open subset of $Y$ which is diffeomorphic to $\Bbb{R}^m$, we see that the restriction of any volume form on $X\times Y$ to $X\times Y_0$ is still a volume form. Hence, $X\times\Bbb{R}^m$ is orientable; let $\mu$ be a volume form here. Now, define $\mu_X:=\iota^*\left(\mu(\cdots, \eta_1,\dots, \eta_m)\right)$, where $\iota:X\to X\times\{0\}\subset X\times\Bbb{R}^m$ is the canonical embedding and for each $i\in\{1,\dots, m\}$, $\eta_i$ is the canonical lift of $\frac{\partial}{\partial y^i}$ on $\Bbb{R}^m$ to $X\times \Bbb{R}^m$. Then, $\mu_X$ is a volume-form on $X$, so $X$ is orientable. Reversing the roles of $X,Y$ in the proof above shows $Y$ is also orientable.
To prove the $(\implies)$ direction, take any timelike vector field $T$ on $B$, and define a vector field $\tau$ on $M=B\times R$ by \begin{align} \tau(b,r):=(T(b),0)\in T_bB\times T_rR\cong T_{(b,r)}(B\times R). \end{align} Then, obviously (suppressing the projections in the notation) we have $g(\tau,\tau)=g_B(T,T)<0$, so $\tau$ is timelike on $M$. Conversely, given any timelike vector field $\tau$ on $M$, we can separate it out into components $\tau(b,r)=(\tau_B(b,r),\tau_R(b,r))\in T_bB\times T_rR$ with $\tau_B,\tau_R$ being smooth vector fields on $M$. Observe that \begin{align} g_B(\tau_B,\tau_B)&=g(\tau,\tau)-\phi^2g_R(\tau_R,\tau_R)\leq g(\tau,\tau)<0, \end{align} where we have used that $g_R$ is Riemannian. So, if we fix $r\in R$ and define $T(b)=\tau_B(b,r)$, then the above calculation shows we have a smooth timelike vector field on $B$.
With these lemmas, it suffices to find 2-dimensional counterexamples, because to get one of higher dimension $n> 2$, we just take a product with an $(n-2)$-dimensional orientable Riemannian manifold (e.g $\Bbb{R}^{n-2}$ with usual positive-definite metric and orientation).
Details for the Mobius band.
We claim the open Mobius band $M$ admits a time-orientable Lorentzian metric, and also one which is not time-orientable. Here are the details:
Let $\Bbb{Z}$ act on $\Bbb{R}^2$ by $n\cdot (x,y):=(x+n,(-1)^ny)$. This is a smooth, free and proper action (exercise), so by the orbit manifold theorem (see Lee’s Introduction to Smooth Manifolds, Theorem 21.10) the quotient $M:= \Bbb{R}^2/\Bbb{Z}$ equipped with the quotient topology carries a unique smooth structure such that the projection $\pi:\Bbb{R}^2\to M$ is a smooth submersion; this is one of the many possible definitions of the Mobius band (if you have some other definition, you can easily show it’s equivalent to this one). Now, let $X=\frac{\partial}{\partial x}$ on $\Bbb{R}^2$; this vector field is in fact $\Bbb{Z}$-equivariant as a map $X:\Bbb{R}^2\to T\Bbb{R}^2$ (relative to the given action on $\Bbb{R}^2$ and the induced tangent action on the tangent bundle), which implies that it descends uniquely to a smooth vector field $\widetilde{X}:M\to TM$ which is $\pi$-related to $X$. Intuitively, since the $x$-direction is simply being wrapped around, rather than being “twisted”, it follows that the $X$ vector field which points along the $x$-axis will have no problems descending to the quotient.
Now, consider the following Lorentzian metrics on $\Bbb{R}^2$: $g_1=-dx^2+dy^2$ and $g_2=dx^2-dy^2=-g_1$. Relative to both of these, the above action of $\Bbb{Z}$ on $\Bbb{R}^2$ is by isometries; and the fibers are discrete, so in particular semi-Riemannian manifolds in their own right, and so $g_1,g_2$ descend to Lorentzian metrics $\widetilde{g_1},\widetilde{g_2}$ on the quotient $M$, such that $g_i=\pi^*\widetilde{g_i}$ (Lee’s Introduction to Riemannian Manifolds, Corollary 2.29 covers the Riemannian case, but with minor modifications you can convince yourself the semi-Riemannian case holds as well). It is easily verified that \begin{align} \pi^*(\widetilde{g_i}(\widetilde{X},\widetilde{X}))=g_i(X,X)=(-1)^i. \end{align} So, for $i=1$, we see (by surjectivity of $\pi$) that $\widetilde{X}$ is a smooth timelike vector field on $(M,\widetilde{g}_1)$, and hence this Lorentzian metric on the Mobius band is time-orientable.
However, for $i=2$, the vector field $\widetilde{X}$ is spacelike, so supposing for the sake of contradiction there existed a smooth timelike vector field $T$ on $M$, it obviously can’t be a multiple of $\widetilde{X}$, and so $\{T,\widetilde{X}\}$ forms a smooth global frame for the tangent bundle. This implies $M$ is orientable as a manifold (since the bivector field $T\wedge\widetilde{X}$ is nowhere-vanishing, and thus by a musical isomorphism, we get a nowhere-vanishing 2-form on $M$), but this is absurd since we know the Mobius band is not an orientable manifold. Therefore, $(M,g_2)$ is not time-orientable.
For the sake of completeness, here’s a proof that $M$ is not an orientable manifold. Supposing it is, let $\mu$ be a volume form on $M$. Then $\pi^*\mu$ is a volume form on $\Bbb{R}^2$ since $\Bbb{Z}$ being discrete implies $\pi$ is a local diffeomorphism. Hence can be written as $f(x,y)\,dx\wedge dy$ for some smooth nowhere-vanishing function $f:\Bbb{R}^2\to\Bbb{R}$, which by the intermediate-value theorem must maintain aconstant sign. But now letting $\theta_n(x,y)=(x+n,(-1)^ny)$ be the action map of the integer $n$, we easily see by pulling back by $\theta_n$ that \begin{align} f\,dx\wedge dy&=\pi^*\mu=(\pi\circ\theta_n)^*\mu=\theta_n^*\pi^*\mu=\theta_n^*(f\,dx\wedge dy)=(-1)^n(f\circ \theta_n)\,dx\wedge dy, \end{align} which implies $f\circ \theta_n=(-1)^nf$, contradicting that $f$ maintains a constant sign (and in fact it says that if we move along the $x$-axis by $n$ units with $n$ odd, then $f$ flips sign, which coincides with our intuitive understanding of the ‘twist’ in the Mobius strip as well).
Details for the Cylinder.
The cylinder $M=S^1\times\Bbb{R}$ is obviously orientable and time-orientable with the metric $g=d\theta^2-dt^2$ (i.e the product of the usual positive-definite round metric on $S^1$ with the negative-definite $(\Bbb{R},-dt^2)$) because $\frac{\partial}{\partial t}$ is well-defined and timelike for $g$, so provides a time-orientation.
To give an example of a non-time-orientable Lorentzian metric on the cylinder, consider the following picture:
The picture on the left is from O’Neill Chapter 5, figure 5 (page 145), and the picture drawn on the right is an example of a vector field in the covering space $\Bbb{R}^2$ of the cylinder. Intuitively, the red vector field rotates clockwise as we right along the $x$-axis, i.e the timecones are themselves rotating; and if we arrange the speed of rotation correctly, the timecones after moving $2\pi$ will be in the flipped order, so when we project to the quotient, we’ll get an inconsistency.
More precisely, on $\Bbb{R}^2$ with coordinates $(x,y)$, let us define the tensor field \begin{align} g&=\cos x\,dx^2-2\sin x\,dx\,dy-\cos x\,dy^2 = -(dx^2+dy^2)+ 2\left(\cos\frac{x}{2}\,dx-\sin\frac{x}{2}\,dy\right)^2, \end{align} where in the second equality I have used the half-angle formulae. Note that the first equality tells us $g$ is symmetric. Next, define the vector fields \begin{align} T=-\sin\left(\frac{x}{2}\right)\frac{\partial}{\partial x} -\cos\left(\frac{x}{2}\right)\frac{\partial}{\partial y},\quad\text{and}\quad X=\cos\left(\frac{x}{2}\right)\frac{\partial}{\partial x} -\sin\left(\frac{x}{2}\right)\frac{\partial}{\partial y}. \end{align} Then, using the second expression for $g$, it is easily seen that $g(T,T)=-1$ and $g(X,X)=1$, and with the first expression $g(T,X)=0$, proving that $g$ has Lorentzian signature and $\{T,X\}$ provides a global $g$-orthonormal frame with $T$ timelike and $X$ spacelike. Btw, the way I got these formulas is I first tried to write down a simple ‘rotating’ vector field like the red arrows above, and this easily gave me the formula for $T$. Then, I started with the general anszatz $g=\alpha\,dx^2+2\beta\,dx\,dy+\gamma\,dy^2$ and tried to choose $\alpha,\beta,\gamma$ to ensure $g(T,T)=-1$, and with a little guess-work I got the above formula, and lastly it was easy to see that $X$ as defined above satisfies $g(X,X)=1$ and $g(T,X)=0$, proving that $g$ is indeed Lorentzian.
Now, consider the action of $\Bbb{Z}$ on $\Bbb{R}^2$ defined as $n\cdot (x,y)\equiv \theta_n(x,y):= (x+2\pi n, y)$. The quotient is the cylinder: $M:=\Bbb{R}^2/_{\theta}\Bbb{Z}\cong (\Bbb{R}/2\pi \Bbb{Z})\times\Bbb{R}\cong S^1\times\Bbb{R}$. Also, the action consists of isometries of $g$: $\theta_n^*g=g$ (using the basic trigonometric facts that $\sin\left(\frac{x}{2}+n\pi\right)=(-1)^n\sin\left(\frac{x}{2}\right)$, and likewise for $\cos$, but this sign disappears because it is squared). Hence, $g$ descends to a Lorentzian metric $\widetilde{g}$ on the cylinder (if by abuse of notation you use local coordinates $(x,y)$ on $S^1\times \Bbb{R}$ with $x$ defined only mod $2\pi$, then it has the same formula as above). Also, it is easy to verify that $\theta_n^*T=(-1)^nT$ (the LHS being pullback of a vector field).
Suppose for the sake of contradiction $(M,\widetilde{g})$ is time-orientable, and let $\widetilde{T}$ be a smooth timelike vector field. Since $\Bbb{Z}$ is discrete, the projection $p:\Bbb{R}^2\to M$ is a covering map, and in particular a local diffeomorphism, so we can define a unique pullback vector field $\tau:=\pi^*\widetilde{T}$ (more precisely, there’s a unique vector field $\tau$ on $\Bbb{R}^2$ which is $p$-related to $\widetilde{T}$). Now, it is easy to verify using $g=p^*\widetilde{g}$ and that $\tau$ is $p$-related to $\widetilde{T}$ to prove that \begin{align} g(\tau,\tau)=p^*\left(\widetilde{g}(\widetilde{T},\widetilde{T})\right)<0, \end{align} and so $\tau$ is timelike. Also, it is straightforward to verify (keeping in mind $p\circ \theta_n=p$) that \begin{align} \theta_n^*\left(g(T,\tau)\right)&=g(\theta_n^*T,\tau)=(-1)^ng(T,\tau). \end{align} Now, since both $T$ an $\tau$ are timelike vector fields it follows $g(T,\tau)$ is nowhere-vanishing hence maintains a constant sign by the intermediate-value theorem. However, this contradicts the above equation which tells us that when $n$ is odd, $g(T,\tau)$ has opposite sign at $(x,y)$ and $\theta_n(x,y)=(x+2\pi n,y)$. Therefore, $(M,\widetilde{g})$ is not time-orientable.
This example specifically is nice in hindsight because the proof really does mimic the intuitive picture, and the contradiction reached via the intermediate-value theorem is a precise way of saying that when we travel $2\pi$ along the $x$-axis, the timecone has flipped.
Some final words about intuition: notice that in both of the examples above, the lack of time-orientability comes from the timecones in the quotient flipping direction as you travel around in a ‘single’ loop (which is only possible due to non simply-connectedness). But a key difference is the reason for the flip. In the Mobius band case, our timecones in the covering space $\Bbb{R}^2$ were the usual ‘upright’ ones, but in the quotient Mobius band, they flipped because the Mobius band itself is defined with a twist. In the cylinder case, the timecones in the covering space $\Bbb{R}^2$ themselves were the ones rotating, and since the quotient is obtained without a twist, it follows that the flip in the covering space still persists in the quotient. In both cases, formalizing the occurrence of the flip is done via the intermediate-value theorem.