Let me develop my remark. It is actually a very nice problem in ``elementary'' differential geometry.
Here is my claim:
Proposition: Let $(v_1,\ldots, v_n)$ be pointwise linearly independent vector fields on an $n$-dimensional manifold $M$. Then the following are equivalent:
$(i)$ For all $i,j$, $$[v_i,v_j] \in \mathcal C^\infty(M) v_i\oplus \mathcal C^{\infty}(M) v_j$$
$(ii)$ Locally, there exist non-vanishing functions $(f_i)_{1\leq i \leq n}$ such that the vector fields $\hat v_i = f_i v_i$ satisfy $[v_i, v_j] = 0$ for all $i,j$.
Remark first that condition $(i)$ is invariant under multiplying each vector field by a function, which shows that $(ii) \Longrightarrow (i)$. For the other direction let us first prove two lemmas:
Lemma 1: Locally, there exists a non-zero function $f_n$ and functions $(b_i)_{1\leq i\leq n-1}$ such that
$$[f_n v_n, v_i] = b_i v_i~.$$
Proof: By property $(i)$, the distribution spanned by $(v_1,\ldots, v_{n-1})$ is integrable. Locally, it is thus given by the kernel of a closed $1$-form $\alpha$.
Since $(v_1,\ldots, v_n)$ are pointwise linearly independent, the function $\alpha(v_n)$ does not vanish. Let us set $f_n = \frac{1}{\alpha(v_n)}$ and $\hat v_n = f_n v_n$. By construction $\alpha(\hat v_n) \equiv 1$.
For all $1\leq i \leq n-1$, write
$$[\hat v_n, v_i] = a_i \hat v_n + b_i v_i.$$
We have
$$a_i = \alpha([\hat v_n, v_i]) = \hat v_n \cdot (\alpha(v_i)) - v_i \cdot (\alpha(\hat v_n)) - \mathrm d \alpha(\hat v_n, v_i) =0$$
since $\alpha(\hat v_n) \equiv 1$, $\alpha(v_i)\equiv 0$ and $\alpha$ is closed. We conclude that $[\hat v_n, v_i] = b_i v_i$. QED
Lemma 2: Locally, there exist positive functions $(f_i)_{1\leq i \leq n-1}$ such that
$$[\hat v_n, f_i v_i] = 0$$
for all $i$.
Proof: This equation is equivalent to
$$\hat v_n \cdot \log (f_i) = -b_i.$$
Let $\phi_t$ denote the flow of $\hat v_n$. Locally, let $L$ be some leaf of the foliation tangent to $(v_1,\ldots, v_{n-1})$. Then every point $y$ in some neighbourhood of $L$ can be written as $\phi_t(x)$ for a unique $t\in \mathbb R$ and $x\in V$. Set
$$f_i(\phi_t(x)) = \exp\left(-\int_{s=0}^t b_i(\phi_s(x))\right)~.$$
Then $f_i$ satisfies the above equation. QED
Proof of the proposition:
We have modified $(v_1,\ldots, v_n)$ into $(\hat v_1, \ldots, \hat v_n)$ such that $[\hat v_n, \hat v_i] = 0$ for all $i$. Now, by induction, we can find local functions $(f_i)_{1\leq i \leq n-1}$ on the leaf $L$ such $[f_i \hat v_i, f_j \hat v_j] = 0$. We extend them to $\phi_t$-invariant functions in a neighbourhood of $V$. Then we still have
$$[\hat v_n, f_i \hat v_i] = 0.$$
Thus
$${\phi_t}_* (f_i \hat v_i) = f_i v_i~,$$
hence
$${\phi_t}_* [f_i \hat v_i, f_j \hat v_j] = [f_i \hat v_i, f_j \hat v_j].$$
Since $[f_i \hat v_i, f_j \hat v_j]$ vanishes on $L$ and is $\phi_t$-invariant, it vanishes in a neighbourhood of $L$. QED
Remark 1: I assumed $n= \dim (M)$ for simplicity, but now, if $n< \dim(M)$ then $v_1,\ldots, v_n$ span the tangent distribution to some foliation, and locally one can modify $v_1,\ldots, v_n$ independently on each leaf so that they commute.
Remark 2: I don't know what happens near points where $v_1,\ldots, v_n$ are not linearly independent anymore. It is quite an interesting question !