Oh Times, $\otimes$ in linear algebra and tensors

If $V$ and $W$ are vector spaces, you can form a third vector space from them called their tensor product $V \otimes W$. The tensor product consists of sums of certain vectors called "pure tensors," which are written $v \otimes w$ where $v \in V, w \in W$, subject to certain rules, e.g. $(v_1 + v_2) \otimes w = v_1 \otimes w + v_2 \otimes w$. For a complete list of these rules see Wikipedia. In practice you'll do fine if you remember the following:

If $v_1, \dots v_n$ is a basis of $V$ and $w_1, \dots w_m$ is a basis of $W$, then the pure tensors $v_i \otimes w_j, 1 \le i \le n, 1 \le j \le m$ form a basis of $V \otimes W$. In particular, $\dim V \otimes W = \dim V \times \dim W$.

If $T : V_1 \to V_2$ and $S : W_1 \to W_2$ are two linear maps, you can form a third linear map from them which is also called their tensor product

$$T \otimes S : V_1 \otimes W_1 \to V_2 \otimes W_2.$$

It is completely determined by how it behaves on pure tensors, which is

$$(T \otimes S)(v \otimes w) = T(v) \otimes S(w).$$

The relationship between these two uses of the term "tensor product" is given formally by the notion of a functor.

Tensor product notation for linear maps is compatible with the notation $v \otimes w$ for pure tensors in the following sense. A vector $v \in V$ in a vector space is the same thing as a linear map $v : 1 \to V$ from the one-dimensional vector space $1$ given by the underlying field to $V$, and if $v : 1 \to V$ and $w : 1 \to W$ are two vectors in $V, W$, then their tensor product as linear maps $v \otimes w : 1 \otimes 1 \to V \otimes W$ corresponds to the pure tensor $v \otimes w$, where we use that there's a canonical isomorphism $1 \otimes 1 \cong 1$.

The Kronecker product is a description of the tensor product of linear maps with respect to a choice of basis for all of the vector spaces involved. Formally, with notation as above, if

1. $B_1, B_2$ are bases for $V_1, V_2$,
2. $C_1, C_2$ are bases for $W_1, W_2$,
3. given bases $B_i, C_i$ of $V_i, W_i$, we write $B_i \otimes C_i$ for the corresponding basis of $V_i \otimes W_i$ as in the highlighted area above, and
4. we write $_{B_2}[T]_{B_1}$ to refer to the matrix of a linear transformation $T : V_1 \to V_2$ with respect to a basis $B_1$ of $V_1$ and a basis $B_2$ of $V_2$,

then we have

$$_{B_2 \otimes C_2}[T \otimes S]_{B_1 \otimes C_1} = \, _{B_2}[T]_{B_1} \otimes \, _{C_2}[S]_{C_1}$$

where on the LHS $\otimes$ means the tensor product of linear maps and on the RHS $\otimes$ means the Kronecker product.

One final remark: the definition of spaces of tensors you give in 2) is a terrible definition that I've only seen in some textbooks on differential geometry. It is absolutely the wrong way to think about tensors.