The existing answer does not directly address the question, « ¿Why? »
For simplicity, consider two distinguishable particles each of whose wave functions would be described by $\cal H_i = L^2(R)$. In order to describe the states of them both considered as a system, we must have at least all the following possibilities:
$\psi_1$, the wave function describing the first particle (say a pion) and $\psi_2$, (say, a deuteron). So the state space we want must have at least all ordered pairs
$(\psi_1,\psi_2)$ in it.
But on the other hand, it must be a Hilbert space. It is a fundamental principle of Quantum Mechanics that physical states can be superposed and you get a new, possible, physical state. So the state space we want must have all formal linear combinations of these ordered pairs in it.
Nevertheless, some of these formal combinations will be physically the same state, for example $(\psi_1 + \phi_1, \psi_2)$ must be the same state of the combined system as $(\psi_1,\psi_2) + (\phi_1,\psi_2)$. Grunging along systematically like this you wind up with the generators and relations for the tensor product of $\cal H_1$ and $\cal H_2$.
Now, there is also a more intuitive way to see this. If $\{ v_i\}$ is an orthonormal basis for the space of the electron, say by energy levels, then since the other particle could be in any state $\psi_2$ independently of what the electron is doing, we have a separate copy of its state space for each $i$. And by superoposition, we must be able to take linear combinations of these different copies since the electron might not have been in an energy eigentstate, but might have been in some superoposition. So we get
$$\sum_i L^2{R_2^3}$$ where I put the subscript 2 to show it is the spatial position of the deuteron. But again, this is just the tensor product.
The third way to see this is the most intuitive and uses the functorial properties of the tensor product. For every wave function $\psi_1(x,y,z)$ for the electron, it describes the probability amplitude for the electron to be found at the point $(x,y,z)$.
Since the particles are distinguishable, each of these possibilities can be associated with a possible state of the deuteron. So to describe the combined system, we should replace $\psi_1$ by a function which, to each point $(x,y,z)$, attaches a possible state of the deuteron, with the probability amplitude of the electron's being at that point. But once again, this is the tensor product since $L^2(V) \otimes L^2(W*)$ is equal to $L^2(L(W,V))$, where the second $L$ is the usual notation for linear maps from one Banach space to another.