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After having read the Wiki article on the Tensor Product, I have tried constructing $\mathbb{R}\otimes\mathbb{R}$ (over the field $\mathbb{R}$), to check my understanding.

Step 1: construct $\mathbb{R}\times\mathbb{R} := \{(a,b); a\in \mathbb{R}, b\in \mathbb{R}\}$. So for example, $\{1,2\},\{0,5\}$ and $\{2,7\}$ are all in $\mathbb{R}\times\mathbb{R}$.

Step 2: Construct $F(\mathbb{R}\times\mathbb{R}):= \{g: \mathbb{R}\times\mathbb{R}\to \mathbb{R}; |\mbox{supp}(g)|<\infty\}.$ So for example, the function $$g(x,y) = \begin{cases} 3 &&\mbox{ if $(x,y)=(1,2)$}\\ 6 &&\mbox{ if $(x,y)=(0,5)$}\\ 1&&\mbox{ if $(x,y)=(2,7)$}\\ 0 &&\mbox{ else } \end{cases}$$ is in $F(\mathbb{R}\times\mathbb{R})$.

Step 3: Construct congruence classes to ensure bilinearity. So for example, one congruence class could be $\{(1,2),(1,1)+(1,1), 2(0.5,2), 2(1,1),...\}$.

This is where I stop understanding. What do the elements of $F(V\times W)$ (such as the function $g$) have to do with these equivalence classes? How are these relations 'on $F(V\times W)$', given that the elements of $F(V\times W)$ are functions?

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    Without the formalism, note that if $a,b\in\mathbb R$, then $a\otimes b=1\otimes ab$. So $\mathbb R\otimes\mathbb R=1\otimes\mathbb R=\mathbb R$. – 2017-02-12

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One obvious basis for $F(V \times W)$ is $\left\{\delta_{(u,w)}, u \in V, w \in W, \right\}$. In wiki they denote those basis elements by $(u,w)$ and define the quotient relations on the basis, extending them by linearity to arbitrary element of $F(V\times W)$. For instance, a function mapping $(1,2)$ to $1$ is identified with function mapping $(1,1)$ to $2$ (in particular $\delta_{(1,2)}=\delta_{(2,1)}$). That can be confusing a little bit but if you think of those delta's as indicator functions it becomes transparent. I think tensor product construction is covered relatively well in Lang's algebra textbook.

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    Ok - so in the example that I gave, $g = 3\delta_{(1,2)}+6\delta_{(0,5)}+\delta_{(2,7)}$ would get identified with, for example, $h = 6\delta_{(1,1)}+6\delta_{(0,0)}+6\delta_{(0,0)}+\delta_{(0,7)}+2\delta_{(1,7)}$, right? – 2017-02-12
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    Yes, in general $\delta_{(0,a)}$ is identified with $\delta_{(b,0)}$ and $\delta_{(0,c)}$. – 2017-02-13
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While it is of course very useful for proving the existence of tensor products in great generality and independent of, e.g, the choice of a basis, the free vector space approach is, in my opinion, not that useful for understanding what the tensor product of two vector spaces actually is.

I think it's much easier to (accept the fact that it exists and can be and is a universal object) and to use known properties of the tensor product to figure out what a certain tensor product actually is.

In your special case ($\mathbb{R}\otimes \mathbb{R}$) one of the most useful facts about tensor spaces is the dimension formula, which tells you that this is a real vector space of dimension $1$, so it's actually $\mathbb{R}$ itself.

Your mileage may vary, of course.