The result is not true and the best upper-bound on $n$ to make it work happens to
be $2m-1$.

Assume that $n<2m$. Let $(t_1,\dots,t_n)$ be a family of real numbers
such that $\sum_{k=1}^n t_k \,x_k \otimes x_k=0$. Assume that some $t_k$ is non-zero.
Denote by N the number of indices $k$ such that $t_k \neq 0$.
Obviously $N>m$. Without loss of generality, we may assume that $t_1,\dots,t_m$ are all non-zero, and we rewrite the above equality as
$\sum_{k=1}^m t_k x_k \otimes x_k=-\sum_{k=m+1}^n t_k x_k \otimes x_k$.
As $(x_1,\dots,x_m)$ and $(x_{m+1},\dots,x_n)$ are both linearly independent,
one sees that the rank of $\sum_{k=1}^m t_k x_k \otimes x_k$ is $m$, whereas the one of
$\sum_{k=m+1}^n t_k\,x_k \otimes x_k$ is at most $n-m<m$. This is a contradiction.

To see that $2m-1$ is an optimal upper-bound, note first that
the problem may be entirely restated in terms of families of symmetric rank $1$ matrices. One takes a family $(X_i)_{1 \leq i \leq n}$ of vectors of $\mathbb{R}^m$
in which every subfamily with $m$ vectors is a basis of $\mathbb{R}^m$,
and one tries to find the rank of the family $(X_i X_i^T)_{1 \leq i \leq n}$ of
rank $1$ symmetric matrices.
Now, for $a \in \mathbb{R}$, consider the vector $X(a)=(a^k)_{0 \leq k \leq n-1}$.
Then, for every $n$-tuple $(a_1,\dots,a_n)$ of real numbers, the vectors
$X(a_1),\dots,X(a_n)$ are linearly independent; however, the matrices $X(a)X(a)^T$
all belong to the space of real $m \times m$ Hankel matrices, which has dimension $2m-1$,
and hence a family consisting of such matrices has rank at most $2m-1$.