If you restrict attention to separable extensions then the picture is as follows. Let $\overline{k}$ be a separable closure of $k$ with Galois group $G$. If $X$ is a finite $G$-set we have a $k$-algebra $A(X)=\text{Map}_G(X,\overline{k})$. This construction gives an equivalence from the category of finite $G$-sets to the category of finite etale $k$-algebras. The finite separable extensions of $k$ just correspond to the $G$-sets on $X$ which $G$ acts transitively (so $X\simeq G/H$ for some $H$, and $A(X)=\overline{k}^H$). We can decompose any $X$ as a disjoint union of transitive $G$-sets, and this writes $A(X)$ as a product of fields. We also have $A(X)\otimes_kA(Y)=A(X\times Y)$. Thus, if $K=\overline{k}^H=A(G/H)$ then we can decompose $K\otimes_kK$ as a product of fields by decomposing $(G/H)\times(G/H)$ as a disjoint union of orbits.

On the other hand, you can consider $k=(\mathbb{Z}/p)(x_1,\dotsc,x_r)$ and define an inseparable field extension $K$ by adjoining $y_i$ with $y_i^p=x_i$ for all $i$. Then $K\otimes_kK$ is generated over $K$ by classes $z_i$ satisfying $z_i^p=x_i$, so the classes $u_i=z_i-y_i$ have $u_i^p=0$. It is not hard to see that $K\otimes_kK$ is just the truncated polynomial algebra generated by $u_1,\dotsc,u_r$ subject to $u_i^p=0$. My guess is that you can't get anything much more complicated than that.

[ADDED LATER]

A finite-dimensional $k$-algebra $A$ is self-injective iff Gorenstein iff $\text{Hom}_k(A,k)$ is isomorphic to $A$ as a $k$-module. There are very many of these. If $(r_1,\dotsc,r_n)$ is any regular sequence in $k[x_1,\dotsc,x_n]$ then the quotient ring $Q=k[x_1,\dotsc,x_n]/(r_1,\dotsc,r_n)$ has this property. The cohomology ring of any closed manifold (with coefficients in $k$) has this property. One small example is $k[x,y]/(x^3,y^2+xy+x^2)$.

If you restrict attention to separable extensions then the picture is as follows. Let $\overline{k}$ be a separable closure of $k$ with Galois group $G$. If $X$ is a finite $G$-set we have a $k$-algebra $A(X)=\text{Map}_G(X,\overline{k})$. This construction gives an equivalence from the category of finite $G$-sets to the category of finite etale $k$-algebras. The finite separable extensions of $k$ just correspond to the $G$-sets on $X$ which $G$ acts transitively (so $X\simeq G/H$ for some $H$, and $A(X)=\overline{k}^H$). We can decompose any $X$ as a disjoint union of transitive $G$-sets, and this writes $A(X)$ as a product of fields. We also have $A(X)\otimes_kA(Y)=A(X\times Y)$. Thus, if $K=\overline{k}^H=A(G/H)$ then we can decompose $K\otimes_kK$ as a product of fields by decomposing $(G/H)\times(G/H)$ as a disjoint union of orbits.

On the other hand, you can consider $k=(\mathbb{Z}/p)(x_1,\dotsc,x_r)$ and define an inseparable field extension $K$ by adjoining $y_i$ with $y_i^p=x_i$ for all $i$. Then $K\otimes_kK$ is generated over $K$ by classes $z_i$ satisfying $z_i^p=x_i$, so the classes $u_i=z_i-y_i$ have $u_i^p=0$. It is not hard to see that $K\otimes_kK$ is just the truncated polynomial algebra generated by $u_1,\dotsc,u_r$ subject to $u_i^p=0$. My guess is that you can't get anything much more complicated than that.

[ADDED LATER]

A commutative finite-dimensional $k$-algebra $A$ is self-injective iff Gorenstein iff $\text{Hom}_k(A,k)$ is isomorphic to $A$ as an $A$-module. There are very many of these. If $(r_1,\dotsc,r_n)$ is any regular sequence in $k[x_1,\dotsc,x_n]$ then the quotient ring $Q=k[x_1,\dotsc,x_n]/(r_1,\dotsc,r_n)$ has this property. The cohomology ring of any closed manifold (with coefficients in $k$) has this property. One small example is $k[x,y]/(x^3,y^2+xy+x^2)$.

If you restrict attention to separable extensions then the picture is as follows. Let $\overline{k}$ be a separable closure of $k$ with Galois group $G$. If $X$ is a finite $G$-set we have a $k$-algebra $A(X)=\text{Map}_G(X,\overline{k})$. This construction gives an equivalence from the category of finite $G$-sets to the category of finite etale $k$-algebras. The finite separable extensions of $k$ just correspond to the $G$-sets on $X$ which $G$ acts transitively (so $X\simeq G/H$ for some $H$, and $A(X)=\overline{k}^H$). We can decompose any $X$ as a disjoint union of transitive $G$-sets, and this writes $A(X)$ as a product of fields. We also have $A(X)\otimes_kA(Y)=A(X\times Y)$. Thus, if $K=\overline{k}^H=A(G/H)$ then we can decompose $K\otimes_kK$ as a product of fields by decomposing $(G/H)\times(G/H)$ as a disjoint union of orbits.

On the other hand, you can consider $k=(\mathbb{Z}/p)(x_1,\dotsc,x_r)$ and define an inseparable field extension $K$ by adjoining $y_i$ with $y_i^p=x_i$ for all $i$. Then $K\otimes_kK$ is generated over $K$ by classes $z_i$ satisfying $z_i^p=x_i$, so the classes $u_i=z_i-y_i$ have $u_i^p=0$. It is not hard to see that $K\otimes_kK$ is just the truncated polynomial algebra generated by $u_1,\dotsc,u_r$ subject to $u_i^p=0$. My guess is that you can't get anything much more complicated than that.

If you restrict attention to separable extensions then the picture is as follows. Let $\overline{k}$ be a separable closure of $k$ with Galois group $G$. If $X$ is a finite $G$-set we have a $k$-algebra $A(X)=\text{Map}_G(X,\overline{k})$. This construction gives an equivalence from the category of finite $G$-sets to the category of finite etale $k$-algebras. The finite separable extensions of $k$ just correspond to the $G$-sets on $X$ which $G$ acts transitively (so $X\simeq G/H$ for some $H$, and $A(X)=\overline{k}^H$). We can decompose any $X$ as a disjoint union of transitive $G$-sets, and this writes $A(X)$ as a product of fields. We also have $A(X)\otimes_kA(Y)=A(X\times Y)$. Thus, if $K=\overline{k}^H=A(G/H)$ then we can decompose $K\otimes_kK$ as a product of fields by decomposing $(G/H)\times(G/H)$ as a disjoint union of orbits.

On the other hand, you can consider $k=(\mathbb{Z}/p)(x_1,\dotsc,x_r)$ and define an inseparable field extension $K$ by adjoining $y_i$ with $y_i^p=x_i$ for all $i$. Then $K\otimes_kK$ is generated over $K$ by classes $z_i$ satisfying $z_i^p=x_i$, so the classes $u_i=z_i-y_i$ have $u_i^p=0$. It is not hard to see that $K\otimes_kK$ is just the truncated polynomial algebra generated by $u_1,\dotsc,u_r$ subject to $u_i^p=0$. My guess is that you can't get anything much more complicated than that.

[ADDED LATER]

A finite-dimensional $k$-algebra $A$ is self-injective iff Gorenstein iff $\text{Hom}_k(A,k)$ is isomorphic to $A$ as a $k$-module. There are very many of these. If $(r_1,\dotsc,r_n)$ is any regular sequence in $k[x_1,\dotsc,x_n]$ then the quotient ring $Q=k[x_1,\dotsc,x_n]/(r_1,\dotsc,r_n)$ has this property. The cohomology ring of any closed manifold (with coefficients in $k$) has this property. One small example is $k[x,y]/(x^3,y^2+xy+x^2)$.