Just some definitions, in case you're unfamiliar with them: Let $\hat{V}$ denote the vector space of linear functions from a vector space $V$ to the scalar field. Remember, a multilinear map is one of the form $V \times V \times \cdots \times V \to W$ (with $n$ copies of $V$), where $W$ is another vector space, such that if we fix $n-1$ of the arguments, the function becomes a linear function from $V$ to $W$ in the argument not fixed. A multilinear form is one in which $W=K$, the scalar field (you can replace $K$ with $\mathbb{R}$ or $\mathbb{C}$ if you like). For example, the inner product on $\mathbb{R}^n$ is a bilinear form on $\mathbb{R}^n$, since if we fix one argument, it becomes linear in the other. If we view an $n \times n$ matrix as a conglomeration of $n$ columns, then the determinant is an $n$-form.

Then $\hat{V} \otimes \hat{V}$ corresponds to the set of bilinear forms, and in general, a tensor product of multiple copies of $\hat{V}$ corresponds to the set of $n$-linear forms (i.e. multilinear forms with $n$ arguments). That, there is a concrete description of tensor products of the dual space with itself, and many books which do not wish to develop the notion of tensor product will use this in place of tensor products. That is, all they must do is define a certain kind of map, and then the tensor product is just the set of maps of that kind. Then how do we explain the tensor product $V \otimes V$ (or more generally $V \otimes U$, where $U$ is another vector space)? We could note that $V$ is canonically isomorphic to its double dual, i.e. the dual space of $\hat{V}$, and then view $V \otimes V$ as the set of bilinear forms on $\hat{V}$. But there is a nicer way, and this uses the universal property.

A bilinear map $V \times V \to W$ corresponds to a linear map $V \otimes V \to W$. If $f(-,-)$ denotes the bilinear map, and $x,y \in V$, then our linear map sends $x \otimes y$ to $f(x,y)$. You could try to think of the tensor product as pairs of vectors, but the tensor product contains elements which are not $x \otimes y$ for some $x,y \in V$. We do have that $x_1 \otimes y_1 + x_2 \otimes y_2$ maps to $f(x_1,y_1)+f(x_2,y_2)$. In more generality, if $W$ and $U$ are two other vector spaces, linear maps $U \otimes V \to W$ correspond to bilinear maps $U \times V \to W$. Then what is an element of $U \otimes V$? It is a thing you stick into a bilinear map. This is the key idea which helped me understand tensor products. I repeat, an element of a tensor product is simply a thing you stick into a bilinear map. In general, elements of some universal construction defined by maps going out of a certain object have some description as "things you stick into some kind of map (or a collection of multiple maps).

Just some definitions, in case you're unfamiliar with them: Let $\hat{V}$ denote the vector space of linear functions from a vector space $V$ to the scalar field. Remember, a multilinear map is one of the form $V \times V \times \cdots \times V \to W$ (with $n$ copies of $V$), where $W$ is another vector space, such that if we fix $n-1$ of the arguments, the function becomes a linear function from $V$ to $W$ in the argument not fixed. A multilinear form is one in which $W=K$, the scalar field (you can replace $K$ with $\mathbb{R}$ or $\mathbb{C}$ if you like). For example, the inner product on $\mathbb{R}^n$ is a bilinear form on $\mathbb{R}^n$, since if we fix one argument, it becomes linear in the other. If we view an $n \times n$ matrix as a conglomeration of $n$ columns, then the determinant is an $n$-form.

Then $\hat{V} \otimes \hat{V}$ corresponds to the set of bilinear forms, and in general, a tensor product of multiple copies of $\hat{V}$ corresponds to the set of $n$-linear forms (i.e. multilinear forms with $n$ arguments). That, there is a concrete description of tensor products of the dual space with itself, and many books which do not wish to develop the notion of tensor product will use this in place of tensor products. That is, all they must do is define a certain kind of map, and then the tensor product is just the set of maps of that kind. Then how do we explain the tensor product $V \otimes V$ (or more generally $V \otimes U$, where $U$ is another vector space)? We could note that $V$ is canonically isomorphic to its double dual, i.e. the dual space of $\hat{V}$, and then view $V \otimes V$ as the set of bilinear forms on $\hat{V}$. But there is a nicer way, and this uses the universal property.

A bilinear map $V \times V \to W$ corresponds to a linear map $V \otimes V \to W$. If $f(-,-)$ denotes the bilinear map, and $x,y \in V$, then our linear map sends $x \otimes y$ to $f(x,y)$. You could try to think of the tensor product as pairs of vectors, but the tensor product contains elements which are not $x \otimes y$ for some $x,y \in V$. We do have that $x_1 \otimes y_1 + x_2 \otimes y_2$ maps to $f(x_1,y_1)+f(x_2,y_2)$. In more generality, if $W$ and $U$ are two other vector spaces, linear maps $U \otimes V \to W$ correspond to bilinear maps $U \times V \to W$. Then what is an element of $U \otimes V$? It is a thing you stick into a bilinear map. This is the key idea which helped me understand tensor products. I repeat, an element of a tensor product is simply a thing you stick into a bilinear map. In general, elements of some universal construction defined by maps going out of a certain object have some description as "things you stick into some kind of map (or a collection of multiple maps)."