Yes, at least for finite dimensional vector spaces. First, I'll show you for two vector spaces $H_1$ and $H_2$ and you can proceed by induction. Use the fact that any linear operator can be represented as a sum of tensor product of linear operators, and if $H$ is Hermitian then $\frac{H + H^\dagger}{2} = H$. That is,
$$H = \frac{\sum_{i = 1}^m A_i \otimes B_i + A_i^\dagger \otimes B_i^\dagger}{2}$$
for linear operators $A_i :H_1 \to H_1$ and $B_i:H_2 \to H_2$. Now
$$ H = \sum_{i = 1}^m \left(\frac{A_i + A_i^\dagger}{2}\right) \otimes \left(\frac{B_i + B_i^\dagger}{2}\right) - \sum_{i = 1}^m \left(\frac{A_i - A_i^\dagger}{2i}\right) \otimes \left(\frac{B_i - B_i^\dagger}{2i}\right),$$
and each term is a tensor product of Hermitian operators.
The concept of tensor products of vector spaces is required for the description of General Relativity. For example, the metric tensor $ \mathbf{g} $ of a pseudo-Riemannian manifold $ (M,\mathbf{g}) $ can be viewed as a ‘disjoint collection’ of tensor products. In terms of a local-coordinate system $ (x^{1},\ldots,x^{n}) $, we can express $ \mathbf{g} $ as $ g_{ij} \cdot d{x^{i}} \otimes d{x^{j}} $, where $ d{x^{i}} $ and $ d{x^{j}} $ are differential $ 1 $-forms and $ g_{ij} $ is just a scalar coefficient. At each $ p \in M $, what $ \mathbf{g} $ does is to take two tangent vectors at $ p $ as input and produce a scalar in $ \mathbb{R} $ as output, all in a linear fashion. In other words, at each point $ p $, we can view $ \mathbf{g} = g_{ij} \cdot d{x^{i}} \otimes d{x^{j}} $ as a bilinear mapping from $ {T_{p}}(M) \times {T_{p}}(M) $ to $ \mathbb{R} $, where $ {T_{p}}(M) $ denotes the tangent space at $ p $.
The metric tensor is the very object that encodes geometrical information about the manifold $ M $. All other useful quantities are constructed from it, such as the Riemann curvature tensor and the Ricci curvature tensor (of course, we still need something called an ‘affine connection’ in order to define these tensors). In General Relativity, the metric tensor encodes the curvature of spacetime, which can and does affect the performance of high-precision instrumentation such as the Global Positioning System (GPS).
For vector spaces over a field $ \mathbb{K} $, tensor products are usually defined in terms of multilinear mappings. You may have seen the more abstract definition using a system of generators, but it can be shown that these two definitions are the same. From the categorical point of view, both constructions satisfy the same universal property (this property is explicated in the Wikipedia article on tensor products), so they must be isomorphic in the category of $ \mathbb{K} $-vector spaces.
You need to understand that tensor products are algebraic objects, while tensor fields are geometrical objects. Even more fundamental than the concept of a tensor field is that of a tensor bundle. You can think of a tensor bundle as a collection of tensor products attached to individual points of a manifold in a consistent manner (by ‘consistent’, I mean that the axioms of a vector bundle must be satisfied). How we relate the tensor product attached to a point to the tensor product attached to another point is what the subject of differential geometry is all about. On the other hand, what we do with the tensor product attached to a fixed single point is pure algebra.
Any discussion of spacetime on a global scale requires General Relativity, which takes into account the curvature of spacetime. However, in a small neighborhood of any fixed point in spacetime, one always has what is called a ‘local inertial frame’. In such a frame, gravitational effects can be transformed away, in which case, General Relativity simply reduces to Special Relativity, which some mathematicians view as nothing more than just advanced linear algebra. This illustration reinforces the idea that ‘global is geometry but local is algebra’.
Best Answer
If you want to study a mathematical object, whether it is a set, manifold, group, vector space, whatever, it is often fruitful to look at natural collections of functions on that space.
Roughly, the purpose of the tensor product, $\otimes$, is to make the following statement true: $$\text{functions}(X \times Y) = \text{functions}(X)\otimes \text{functions}(Y)$$
The specific details about which spaces of functions to choose depend on the type of mathematical object you are interested in.
Here's a pdf that explains it better than I can, http://abel.math.harvard.edu/archive/25b_spring_05/tensor.pdf