I think many of the answers assume you know what tensor product is...

A particularly illuminating example might help here. Before I give it, I'd like to mention that mathematicians don't acknowledge the origin of the name tensor. The meaning was probably lost in over use, but is preserved in translations in other languages (I'm Chinese). Tensor looks like tension, and (I imagine) was used first to describe tension of a membrane or something. (Then Riemann took it up as something that has more than two indices, for his curvature for example. I could be wrong about the history.)

If you want to tensor a plane with another plane, that's not gonna be particularly illuminating. It'll be a four-dimensional space, but one hardly sees what it has to do with the two planes.

Let $V$ be the vector space of polynomials of degree $\leq n$, and let $W$ be the vector space of polynomials of degree $\leq m$. As you know, the tensor product $V\otimes W$ should consist of all things of the form $v\otimes w$, for each $v\in V$ and each $w\in W$, and all (finite) linear combinations of it. Luckily in this case, we have a good candidate for what $v\otimes w$ is (or is being identified with). For $v=p(x)$ a polynomial in $x$, and $w=q(y)$ a polynomial in (another variable) $y$, then $v\otimes w$ is simply the product $p(x)q(y)$ as a polynomial of two variables. Now you see what the tensor product is: $V\otimes W$ is the vector space of polynomials of two variables, whose degree in $x$ is at most $n$, and degree in $y$ is at most $m$. Obviously $\dim V\otimes W = \dim V\times\dim W$.

It might be worthwhile to think about what happens to $V\otimes V$ when you make a change of basis on $V$.

It would be a sin not to mention the extension in infinite-dimensional case. $k[x]\otimes k[y]=k[x,y]$, as you can check using the universal property.

I hope it helps.