The point of this answer is to flesh out my comment above: Any such $X_i$ must be everywhere linearly independent, so they only exist if the manifold $M$ is parallelizable. Suppose, for the sake of contradiction, that the $X_i$ become linearly dependent at some point $p$. Pass to local coordinates $(x_1, \ldots, x_n)$ with $p=(0,0,\ldots, 0)$ and write
$$X_i = \sum_j a_{ij}(x) \frac{\partial}{\partial x_j}.$$
Let $A(x)$ denote the matrix $(A_{ij}(x))$. Our hypothesis is that $A(0)$ is not of full rank.

We compute the [principal symbol][1] of $\sum (\partial/\partial X_i)^2$. We have 
$$\left( \sum_j a_{ij}(x) \frac{\partial}{\partial x_j} \right)^2  = \sum_{j_1, j_2} a_{i j_1}(x) a_{i j_2}(x) \frac{\partial^2}{(\partial x_{j_1}) (\partial x_{j_2})} + \mbox{first order operators}.$$
So the principal symbol is
$$\sum_i \sum_{j_1, j_2} a_{i j_1}(x) a_{i j_2}(x) y_{j_1} y_{j_2} = y^T A^T(x) A(x) y$$
where $y$ is the vector $(y_1, y_2, \ldots, y_n)^T$. If $A$ is not of full rank, neither is this quadratic form.

But the symbol of $\Delta$ is positive definite, a contradiction.


  [1]: https://mathoverflow.net/questions/3477/what-is-the-symbol-of-a-differential-operator