I have $K_1$ and $K_2$ two finite extensions of $\Bbb Q$. I can construct $K_1 \otimes_\Bbb Q K_2$.
This is clearly isomorphic to a direct sum of field as vector space (indeed one can easily see that it is isomorphic to a direct sum of $n_1 \times n_2$ copies of $\Bbb Q$ where $n_i$ is the finite dimension of $K_i$ over $\Bbb Q$ simply by writing a base of $K_1 \otimes_\Bbb Q K_2$ from the bases of $K_1$ and $K_2$).
Now I would like to know if this is still true as rings. i.e. is there a ring isomorphism from the ring $K_1 \otimes_\Bbb Q K_2$ to a direct sum of field.
Yes, this is true. Using the primitive element theorem, you can write $K_1 = \mathbb{Q}[x]/f(x)$ for some irreducible polynomial $f$. Then
$$K_1 \otimes K_2 \cong K_2[x]/f(x).$$
Since $K_2$ is separable, so is the polynomial $f(x)$, so we get a product of fields corresponding to the irreducible factors of $f(x)$ over $K_2$.
This is not true if $\mathbb{Q}$ is replaced by a field which is not perfect. For example, if $k = \mathbb{F}_p(a)$ and $K_1 = K_2 = k[x]/(x^p - a)$, then
$$K_1 \otimes K_2 \cong K_1[x]/(x^p - a) \cong K_1[x]/(x - \sqrt[p]{a})^p$$
is not reduced. In this case it's necessary to assume additionally that $K_1$ and $K_2$ are separable.