$(Y-a) g = 0 \in k(a)[Y]/(f)$ $(a'-a) g(a') = 0 \in k(a)(a')$ This is hard to convert to $k(a) \otimes k(a)$ explicitly because $g$ has both $a$ and $a'$
So let's work with an example: $K = \Bbb Q(\sqrt2)$, $k = \Bbb Q$
then $f = X^2 - 2 = (X-\sqrt2)(X+\sqrt2)$
replace $\sqrt2$ with $\sqrt 2 \otimes 1$ and $X$ with $1 \otimes \sqrt2$
I don't understand why k[X,Y]/(f,f) is the same as k(a)[X]/(f)
But if that's true I understand the rest of the argument
Here's my argument
1) First our ring is not a field, because the product $K \otimes_k K \to K$ is a ring map; maps of fields are injective; this contradicts the dimension count.
2) Artinian domains are fields. For consider the decreasing sequence $(a^n)$ for nonzero $a$. This must stabilize, so for some $m$ we have $(a^m) = (a^{m+1})$. So $a^m = a^{m+1} y$ for some $y$.
Because the ring is a domain, $a^m$ is not zero. Therefore $y$ is not zero. So we have $a^m(ay-1) = 0$. We have constructed an inverse to an arbitrary nonzero $a$.
Finally $K \otimes_k K$ is Artinian, as ideals are $k$-subspaces and this is a finite dimensional $k$-vector space.
@LeakyNun yeah clearly, thanks. i guess i got worried that we might forcibly add new roots in this extension but it's not a splitting field for that polynomial