I've seen a couple times on MathOverflow that $L \otimes_K L \cong L^{\oplus n}$, where $n$ is the degree of $L$ over $K$. Unfortunately, neither of the links has a proof of this fact (most of the discussion, which is way over my head, seems to just take this for granted) and I don't have much intuition for why this should be true.
In particular, this implies that $L \otimes_K L$ isn't even a field, which is surprising -- I'd just think that the inverse of $a \otimes b$ would be $\frac{1}{a} \otimes \frac{1}{b}$. The only time this wouldn't exist is if one of $a$ or $b$ is $0$, but then $a \otimes b$ would be $0$, so this isn't a problem; what's wrong with this explanation?
It would also be nice if someone could actually prove the isomorphism in a way that's understandable to me; my background is 3 semesters of algebra.
Let's look at the most down to earth example: $\mathbb{C} \otimes_\mathbb{R} \mathbb{C}$.
Viewed as an $\mathbb{R}$ vector space it is 4-dimensional, hence strictly bigger than $\mathbb{C}$, yet still finite dimensional. Also it contains $\mathbb{C}$, so if it is a field, it is a finite dimensional field extension of $\mathbb{C}$. This seems weirdly at odds with the fact that $\mathbb{C}$ is algebraically closed - normally you get field extensions by adding roots of polynomials that didn't have roots in the field already. Something gotta give. But what?
Following up on the algebraically closed idea, let's look at the roots of the equations $x^2 = -1$ in $\mathbb{C} \otimes_\mathbb{R} \mathbb{C}$. I can list already three, perhaps there are more:
$I_1 = i \otimes 1$, $I_2 = 1 \otimes i$, $I_3 = -i \otimes 1$ etc. Again we feel that something fishy is going on: uniqueness of factorization in the ring of polynomials would tell us that there can be only two solutions of $x^2 = -1$. Where is our mistake?
Before we go on let's check that $I_1$ and $I_2$ are indeed square roots of $-1$. We have $I_1^2 = -1 \otimes 1$ and $I_2^2 = 1 \otimes -1$.
The reason that these are equal to each other is that $-1$ is an element of $\mathbb{R}$ (the $K$ in your question) and hence we can move it freely over the $\otimes$-sign.
The reason that $-1 \otimes 1 = 1 \otimes -1$ is equal to the element $-1$ in the ring $\mathbb{C} \otimes_\mathbb{R} \mathbb{C}$ is that it can also be written as the product of the real scalar $-1$ with the ring element $1 \otimes 1$ and it is obvious that the latter plays the role of $1$ in the ring $\mathbb{C} \otimes_\mathbb{R} \mathbb{C}$.
Ok, now that we all agree that $I_1^2 = I_2^2$ we need to establish something else: that they are not real scalar multiples of each other and hence linear independent elements of the real vector space $\mathbb{C} \otimes_\mathbb{R} \mathbb{C}$. I leave that to you.
Now look at the product $(I_1 + I_2)(I_1 - I_2)$. By what we just said these are two non-zero element of $\mathbb{C} \otimes_\mathbb{R} \mathbb{C}$ that multiply to $I_1^2 - I_2^2 = 0$. (The actual value of $I_1^2 = I_2^2$ is irrelevant here.)
In fields you can't have non-zero elements multiplying to zero, so $\mathbb{C} \otimes_\mathbb{R} \mathbb{C}$ is not a field.
One thing about this example I want to stress is this: both $I_1 + I_2$ and $I_1 - I_2$ are not pure tensors (i.e. elements of the form $a \otimes b$), but sums of multiple (in this case: 2) pure tensors. Tensor products of vector spaces are full of elements like this, but they tend to be overlooked.