Let $X_1,\dots,X_n \in\mathbb{R}^m$ be vectors; and $M\in\mathbb{R}^{m\times m}$ is a symmetric matrix. My question is as follows. Is there any condition $\mathcal{C}$ on $X_1,\dots,X_n$ such that, under $\mathcal{C}$, it holds: $$ X_i^T M X_i = 0 \iff M=0, $$ and in the absence of $\mathcal{C}$, $X_i^T MX_i = 0\iff M=0$ fails, that is, there is a matrix $M\neq 0$ such that $X_i^T M X_i= 0$ but $M\neq 0$.
For instance, if ${\rm span}(X_iX_i^T)$ is the set of all ($m\times m$) symmetric matrices, then one can establish $X_i^T M X_i = 0\iff M=0$.
The condition that the matrices $X_iX_i^T$ span the symmetric matrices is both sufficient (as you indicated) and necessary. That is, the condition that $X_iX_i^T$ span the symmetric matrices will work as your $\mathcal C$.
Note that the map $\langle A,B \rangle = \operatorname{tr}(AB)$ forms an inner product over the space $\mathcal S$ of symmetric matrices. Now, suppose that $\{X_iX_i^T\}$ fail to span $\mathcal S$. It follows that there exists a non-zero symmetric matrix $M$ in the orthogonal complement $\operatorname{span}(\{X_iX_i^T\})^\perp$ (relative to $\mathcal S$). Although $M$ is non-zero, we find that for any of our $X_i$ we have $$ X_i^TMX_i = \operatorname{tr}(X_i^TMX_i) = \operatorname{tr}(MX_iX_i^T) = \langle M, X_iX_i^T \rangle = 0. $$