I am just going to answer the first question. For a two-tensor we have
$$|T|^2 = \left< T,T \right> = g^{ik}g^{jl}T_{ij}T_{kl},$$
where $g$ is the metric, and the summation convention is understood. Note that this is the standard inner product structure induced by $g$, as mentioned by Jason DeVito. You might like to try and prove that this really does define a norm in the Riemannian case.
More generally, for a $(k,l)$ ($k$ times contravariant, $l$ times covariant) tensor field $T$ (on the manifold) we have
$$|T|^2 = \left< T,T \right> = g^{j_1q_1}g^{j_2q_2}\cdots g^{j_lq_l}g_{i_1p_1}g_{i_2p_2}\cdots g_{i_kp_k}T^{i_1i_2\cdots i_k}_{j_1j_2\cdots j_l}T^{p_1p_2\cdots p_k}_{q_1q_2\cdots q_l}.$$
For your other questions, you could do worse than look in 'Einstein Manifolds' by Besse.
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.)
Best Answer
In general, no, the only parallel symmetric $2$-tensors on a (connected) Riemannian manifold $(M, g)$ are scalar multiples of the metric: Given any parallel, symmetric $2$-tensor $A$ that is not a multiple of $g$ and any point $x \in M$, the holonomy group $\operatorname{Hol}_x(g)$ of $g$ at $x$ is contained in the stabilizer of $A_x$ in $O(g_x)$, which is a positive-codimension subgroup, but the holonomy group of a generic metric is $O(n, \Bbb R)$ or $SO(n, \Bbb R)$.
Since positive definiteness of a symmetric $2$-tensor is an open condition, if $A$ a nonzero, parallel, symmetric $2$-tensor, then at least for small $\epsilon$, $g + \epsilon A$ is also a metric parallel with respect to the Levi-Civita connection $\nabla^g$ of $g$, and by uniqueness, $\nabla^g$ is also the Levi-Civita connection of $g + \epsilon A$: We say that $g, g + \epsilon A$ are affinely equivalent.
A theorem of Eisenhart describes (locally) exactly when this happens (see p. 303 of the reference): If $A$ is a parallel symmetric $2$-tensor on $(M, g)$, we can write it locally as $A = \sum_i \alpha^i g_i$ for some local decomposition $(U, g\vert_U) = (U_1, g_1) \times \cdots \times (U_n, g_n)$ of $(M, g)$ into a Riemannian product (necessarily the coefficients $\alpha^i$ are constants). In particular, if $(M, g)$ does not locally admit a Riemannian product decomposition around some point, $A$ must be a multiple of $g$.
Conversely, we can use this description to give simple examples. For any nontrivial product manifold $(M, g) \times (N, h)$, every symmetric $2$-tensor of the form $\alpha g \oplus \beta h$ is parallel, but among these tensors only those for which $\alpha = \beta$ are multiples of the product metric $g \oplus h$.
One can say more. See: