Having three bilinear, symmetric and positive definite, i.e. inner products, forms it is not always possible to find the basis of the space which is orthogonal simultaneously for all of them. For example take the forms given in the standard basis of $\mathbb{R}^2$:
$$\begin{bmatrix}1&0 \\\ 0&1\end{bmatrix}, \begin{bmatrix}1&0 \\\ 0&2\end{bmatrix}, \begin{bmatrix}4&1 \\\ 1&8\end{bmatrix}.$$
Are there any reasonable and non trivial assumption on the set of three, or any $k$, forms to be simultaneously diagonalizable in some basis?
You are asking whether there is an invertible matrix $A$ such that $$(AS_1A^t,\ldots, AS_kA^t)=(D_1,\ldots,D_k),$$ where $S_1,\ldots, S_k$ are symmetric and positive definite and $D_1,\ldots, D_k$ are positive diagonals. If such $A$ exists we say that $A$ solves your problem for $S_1,\ldots, S_k$.
First notice that $A$ solves your problem for $S_1,\ldots, S_k$ if and only if $AS_1^{\frac{1}{2}}$ solves your problem for $Id, B_2,\ldots, B_k$, where $B_i=S_1^{-\frac{1}{2}}S_iS_1^{-\frac{1}{2}}$.
Now, if $M$ solves your problem for $Id, B_2,\ldots, B_k$ then $M(Id)M^t=D_1$. Hence $D_1^{-\frac{1}{2}}MM^tD_1^{-\frac{1}{2}}=Id.$ Let $O=D_1^{-\frac{1}{2}}M$. So $O$ is orthogonal, since $OO^t=Id$.
In addition, $OB_iO^t=D_1^{-\frac{1}{2}}MB_iM^tD_1^{-\frac{1}{2}}$ are also diagonal positive matrices, since $MB_iM^t$ are.
So the orthogonal matrix $O$ solves your problem for $Id, B_2,\ldots, B_k$, but this means that there is a common orthonormal basis of eigenvectors for the symmetric matrices $Id, B_2,\ldots, B_k$, which is true if and only if they commute.
So the criterion you are looking for is: From $S_1,\ldots, S_k$ find the corresponding $Id, B_2,\ldots, B_k$ then determine whether these new matrices commute.