Let $M$ be an n-dimensional manifold endowed with a Riemannian metric. Suppose we have a coordinate chart, say $(U,\varphi)$ where $U\subset M$ and $\varphi\colon U\rightarrow {\mathbb R}^n$, and let $\partial\_1,\ldots\partial\_n$ denote the corresponding coordinate vector fields. Suppose $\partial\_n$ is bounded. 
Is it possible to construct a new coordinate chart $(U,\varphi')$ (same $U$) in such a way that the following two conditions:


1. At each point the subspace generated by $\{\partial\_1',\ldots,\partial\_{n-1}'\}$ is the same as that generated by $\{\partial\_1,\ldots,\partial\_{n-1}\}$.
2. $||\partial\_n'||=1$ at all points,


are satisfied?


I have no idea. Condition 2 seems to be a little strong. 



Coordinates on Riemannian manifolds
Not always. 


Let $M = \mathbb{R}^2 \setminus \mathbb{R}^+$ with the induced metric (plane with a slit). Let $U = M$ and $\varphi$ be the polar coordinate map $\varphi = (r,\theta)$. Let $r = x^1$ and $\theta = x^2$.


Your condition 1 requires that the level sets of $x^2$ in the primed coordinate system be the same as the level sets of $\theta$. But this means regardless of what you choose as the coordinate function $x^1$, you must have $\|\partial\_2'\| \to \infty$ as "$r\to \infty$", due to the infinite separation of leaves of $\theta$ near infinity. 




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Observing that defining the distribution of $\{ \partial\_1, \ldots, \partial\_{n-1}\}$ is in fact equivalent to defining the level sets of the coordinate $x^n$ (modulo topological concerns), you can rephrase your question as the following:


Give a Riemannian manifold $U$ with metric $g$, and a non-degenerate function $v:U\to\mathbb{R}$ (in the sense that $\mathrm{d}v \neq 0$), can you find a vector field $\eta$ on $U$ such that $g(\eta,\eta) = 1$ and $\eta(v) = 1$. 


The objection above the fold comes from the minmax characterisation:


$$ \|\mathrm{d}v\| = \max\_{\eta: \|\eta\| = 1} \eta(v) $$


so that if $\|\mathrm{d}v\| < 1$ the problem cannot be solved; now, the $v$ should be a function with the same level sets as $x^n$, so if $\|\mathrm{d}x^n\|$ is not bounded below (from zero) initially, there is no hope of rescaling $v$ to satisfy the characterization. 


Now, what if $\mathrm{d}v$ is bounded from below? by rescaling we can assume that it is bounded from below by 1. Next, since in the original question the level sets of $v$ are coordinated, we can assume that there exists also a non-vanishing vector field $\tau$ such that $\tau(v) = 0$ on our manifold $U$. Since $\tau$ is non-vanishing we can further assume that $g(\tau,\tau) = 1$. 


Then to construct the vector field $\eta$ we can take the ansatz


$$ \eta = \frac{1}{\|\mathrm{d}v\|^2}\mathrm{d}v^\sharp + \alpha \tau $$


But construction $\eta(v) = 1$. And we just need to solve for $\alpha$ using the *algebraic* equation


$$ \|\eta\| = 1 \iff \frac{1}{\|\mathrm{d}v\|^2} + \alpha^2 = 1 \iff \alpha = \pm \sqrt{ 1 - \frac{1}{\|\mathrm{d}v\|^2}} $$


