Kostrikin's Definition of Tensor Product

For simplicity, I will only talk about the construction of the tensor product of two $\mathbb{K}$-spaces, $L_1$ and $L_2$. The generalization to the tensor product of any finite number of spaces follows easily.

The intuition is this: Whatever the tensor product $L_1 \otimes L_2$ turns out to be, I want it to be generated by elements of the form $l_1 \otimes l_2$ with $l_1 \in L_1$ and $l_2 \in L_2$. I also want the canonical map $L_1 \times L_2 \to L_1 \otimes L_2$ with $(l_1,l_2) \mapsto l_1 \otimes l_2$ to be $\mathbb{K}$-bilinear, so that means I need the generators of $L_1 \otimes L_2$ to satisfy the following relations:

$$al_1 \otimes l_2 = a(l_1 \otimes l_2),$$ $$l_1 \otimes al_2 = a(l_1 \otimes l_2),$$ $$(l_1 + l_1')\otimes l_2 = l_1\otimes l_2 +l_1' \otimes l_2,$$ $$l_1 \otimes (l_2+l_2') = l_1 \otimes l_2 + l_1 \otimes l_2'.$$

And this is ALL that I want! Nothing more! I want $L_1 \otimes L_2$ to satisfy these properties, and I want absolutely no other relations (besides of course the relations that come from the definition of a vector space). The end product is a $\mathbb{K}$-space $L_1 \otimes L_2$ consisting of $\mathbb{K}$-linear combinations of the generators, and these generators satisfy the above relations.

The fancy language in the technical definition is just a means of formalizing this type of construction (the tensor product can be defined using its universal mapping property, but then one still has to prove that such a gadget actually exists). In fact, this type of construction has a name: We are defining $L_1 \otimes L_2$ by "generators and relations." Everything following this point in my answer is just formalization (keep the above intuition in mind).

The first thing I do is start out with my generators. So in part (1), I start with the $\mathbb{K}$-space $M$ with basis $L_1 \times L_2$. What is this? Formally, it consists of all functions $L_1 \times L_2 \to \mathbb{K}$ with finite support (and the vector space operations are pointwise). Why are those $\delta$'s actually a basis? Take an arbitrary element $m\in M$ (it's a finitely-supported function $L_1 \times L_2 \to \mathbb{K}$), and say it's nonzero at $(l_1,l_1'),\ldots,(l_n,l_n')$. Say $m(l_k,l_k')=a_k$. Then take a moment to convince yourself that the expression

$$ m = \sum_{k=1}^n a_k \delta(l_k,l_k')$$

is unique (remember the operations in $M$ are pointwise). This proves the $\delta$'s are a basis. I wanted $M$ to be the $\mathbb{K}$-space with basis $L_1\times L_2$, and $\delta(l_1,l_2)$ is how I think of $(l_1,l_2) \in L_1 \times L_2$ as living in $M$. Nobody ever thinks of these types of vector spaces (or modules) by their formal definition, we just think of them as all (formal) finite linear combinations of the generators. That is what is happening in (1) when he drops the $\delta$ notation.

Now that I have all my generators in order, I want to introduce the above relations ((2) in your definition). Ultimately, I want, for instance, $(al_1,l_2) = a(l_1,l_2)$ to be true. That is the same as wanting $(al_1,l_2) - a(l_1,l_2) = 0$ to be true. Hence, I want to kill the element $(al_1,l_2) - a(l_1,l_2)$ in $M$. How do you kill elements in vector spaces and modules? Mod out by them. So I take all of the elements that represent my relations, and consider the space $M/M_0$ where $M_0$ is the subspace generated by all of the elements representing my relations. Now the relation $(al_1,l_2) = a(l_1,l_2)$ really is true (modulo $M_0$) in the space $M/M_0$.

Now I set $L_1 \otimes L_2 = M/M_0$ and write $l_1 \otimes l_2$ for the image of $(l_1,l_2)$ in $L_1 \otimes L_2$. As mentioned above, this process completely formalizes my above "working definition" for the tensor product. As Atiyah and MacDonald say in their commutative algebra text, having seen this construction, you may safely forget it. The important part about the tensor product are its mapping properties.

And about those mapping properties... The tensor product, as you probably know, enjoys the following property: Any $\mathbb{K}$-bilinear map $L_1 \times L_2 \to V$ ($V$ a vector space) uniquely factors through the tensor product. To prove that our above definition satisfies this property, we use a "composition" of two universal mapping properties. The process of forming the vector space with a given basis has a universal mapping property (see free module, it's essentially "extending by linearity"), and quotients have their own universal mapping property. This last property is the following: If $V$ is a vector space and $V_0$ is a subspace, then any linear map $f: V \to W$ to another space $W$ which kills $V_0$ uniquely factors through the quotient $V/V_0$. Notice that we used both of these gadgets in our construction of $L_1 \otimes L_2$; if we put these two universal properties together, we obtain the universal mapping property of the tensor product.