Suppose you have a pair of elements in vector spaces, $v\in V$ and $w\in W$. Now suppose that at some future point you're going to compute $f(v,w)$ where $f$ is a bilinear function. For example, when $V=W=\mathbb{R}^3$, $f$ might be the familiar dot or cross product. But it might be something else entirely. In fact, suppose you don't know in advance what $f$ is going to be.
The tensor product precisely answers the question "what information do I need about $v$ and $w$ in order to be able to compute $f(v,w)$ at some future point, whatever $f$ turns out to be?" You could say "knowing $v$ and $w$ is enough information". But that that's more information than you need. $v\otimes w$ contains less information than $(v,w)$ and actually contains the least information you can get away with, and still be able to compute $f(v,w)$ for any bilinear $f$.
I don't know if this will be helpful for you, but when I realised this everything suddenly became crystal clear. It's really just a restatement of the universal property.
