Let $M$ be a smooth manifold of dimension $n$, let $m < n$, $x_1, \ldots, x_m: U \to \mathbb R$, where $U$ is an open subset of $M$, be smooth functions such that $dx_1, \ldots, dx_n$ are linearly independent at $(T_pM)^{\ast}$ for each $p \in U$, then $x_1, \ldots, x_m$ can be completed to a chart of $M$.
The idea seems straightforward, just take a basis of $(T_pM)^{\ast}$, say $dx_1,\ldots, dx_m, y_{m+1}, \ldots, y_{n}$, then the idea is to locate functions $f_{m+1}, \ldots, f_{m+n}$ such that $df_{m+1} = y_{m+1}, \ldots, df_{n} = y_{n}$. How should I do this?
Some Clarification: The new chart is not necessarily required to take values on the whole $U$. The idea is that if we can find functions $df_{m+1} = y_{m+1}, \ldots, df_{m+n} = y_{m+n}$, then by inverse function theorem, in a suitable neighborhood, we can form a chart.
Update: This statement comes from the notes of Prof.Alexy Zinger(corollary 4.12 on page23)
Let me try to answer this question. Given $V_p \neq 0$, a covector in $(T_pM)^{\ast}$, I claim that I can cook up a smooth function $f: U \mathbb \to \mathbb R$, where $U$ is an open neighborhood of $p$, such that $(df)_p = V_p$. Take the dual vector $W_p$ of $V_p$ in $T_pM$, can extend it to a vector field $W$ on $M$. Take its integral curve $\gamma:(-\epsilon, \epsilon)$, such that $\gamma(0) = p, \dot \gamma(0) = W_p$. Then by inverse function theorem $\gamma$ has an inverse in around a neighborhood of $0$, say $\gamma^{-1}$. Notice that $d\gamma^{-1}_{p}\gamma_{0}= d\gamma^{-1}_p d\gamma_{0} = Id_{T_0(\text{some open subset of} (-\epsilon, \epsilon))}$, which implies that $d\gamma^{-1}_pd\gamma_{0}(\frac{\partial}{\partial t}|_{0}) = d\gamma_p^{-1}(W_p) = 1$. So $d\gamma_p = V_p$.