$\lvert A\rangle \langle B \rvert$ is the tensor of a ket and a bra (well, duh). This means it is an element of the tensor product of a Hilbert space $H_1$ (that's where the kets live) and of a dual of a Hilbert space $H_2^\ast$, which is where the bras live. Although for Hilbert spaces their duals are isomorphic to the original space, this distinction should be kept in mind. So we can "feed" a ket $\lvert \psi\rangle$ from $H_2$ to the bra in $\lvert \phi\rangle\otimes \langle\chi\rvert \in H_1\otimes H_2^\ast$, and are left with a state in $H_1$ given by $\langle \chi \vert \psi\rangle \lvert \phi\rangle$. The usual use case for such a tensor product is when $H_1=H_2$ to construct a map from $H_1$ to itself, e.g. the projector onto a state $\lvert \psi \rangle$ is given by $\lvert\psi\rangle \langle \psi \rvert$.
In general, a tensor in $H_2 \otimes H_1^\ast$ corresponds to a linear operator $H_1\to H_2$. In the finite-dimensional case, these are all linear operators, in the infinte-dimensional case, this is no longer true, e.g. $H^\ast \otimes H$ are precisely the Hilbert-Schmidt operators on $H$.
In constract, a tensor $\lvert A\rangle\otimes \lvert B\rangle$ (also just written $\lvert A \rangle \lvert B\rangle$) in $H_1\otimes H_2$, although it corresponds to a bilinear map $H_1\times H_2\to\mathbb{C}$ by definition, is usually not meant to denote an operator, but a state. Given two quantum systems $H_1$ and $H_2$, $H_1\otimes H_2$ is the space of the states of the combined system (as for why, see this question).
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Difference between Cartesian and tensor product
When the Cartesian product is equipped with the "natural" vector space structure, it's usually called the direct sum and denoted by the symbol $\oplus$. As other answers state, the direct sum (Cartesian product) and the tensor product of two vector spaces can be clearly seen to be different by their dimension.
If $\{v_i\}$ and $\{w_i\}$ are basis of $V$ and $W$, we have that $\{v_i\}\cup\{w_j\}$ is a basis of $V\oplus W$ and $\{v_i\otimes w_j\}$ is a basis of $V\otimes W$. Therefore,
$$\operatorname{dim}(V\oplus W)=\operatorname{dim}V+\operatorname{dim}W$$ $$\operatorname{dim}(V\otimes W)= \operatorname{dim}V\cdot\operatorname{dim}W$$
As you can see, the "Cartesian product" behaves more like a sum when dealing with vector spaces whereas the role of a product is adopted by the tensor product.
Composite systems
Now, to get some intuition about why we should use the tensor product of the spaces of states of two different quantum systems when we want to describe the composite system we can use the following analogy.
Consider two classical systems with a finite number $m$ and $n$ of states $\{s_i\}$ and $\{r_i\}$. When describing the joint system, would we want to have as space of states the union or the Cartesian product of the original sets?
We would like the product, because we expect to have all the possible combined states $\{S_{ij}=(s_i,r_j)\}$ for all $i$, $j$. As you can see, number of elements of the new set is $m\cdot n$.
In the quantum analog of this setting, the two subsystems are described by vector spaces whose basis are $\{s_i\}$ and $\{r_i\}$. In the same way, the composite system will have a basis $\{S_{ij}= (s_i,r_j)\}$ (which we write as $S_{ij}=s_i\otimes r_j$) so it must be the tensor product.
Note: The key of this argument is the observation that the set whose number of elements is the product is the Cartesian product (classical case), whereas the vector space whose dimension is the product is the tensor product (quantum case).