I've stumbled upon an ostensible inconsistency which probably has a simple resolution I'm overlooking.
Consider for the time being a smooth function $f: \mathbb{R} \to \mathbb{R}$ giving rise to a manifold $M = \mathrm{graph}(f)\subset \mathbb{R}^{2}$. Introducing the two (global) orthonormal unit vectors $e_{1}$ and $e_{2}$ of $\mathbb{R}^{2}$, a chart $(U, \varphi)$ of $M$ is given with $$\varphi(u^{1}) = x^{1}(u^{1})e_{1} + x^{2}(u^{1})e_{2} = u^{1}e_{1} + f(u^{1})e_{2},$$ where it's fair to say that $x^{1} = u^{1}$ holds. Let the local tangent vector at point $p = (p_{1}, p_{2})\in M$ now be given by $$\frac{\partial}{\partial u^{1}}\bigg \vert_{p} = e_{1} + f'(p_{1})e_{2}.$$
My question is with respect to the corresponding differentials of the coordinates, $\mathrm{d}x^{1}$ and $\mathrm{d}u^{1}$, the former of which, for a given point $p\in M$, acts on the tangent space $T_{p}\mathbb{R}^{2}$, which happens to be $\mathbb{R}^{2}$. By contrast, the latter acts on the tangent space $T_{p}M$, which only consists of the linear span of $e_{1} + f'(p_{1})e_{2}$. Thus, their domains do not coincide. In addition, both quantities differ with respect to which tangent vectors ($e_{1} \mathrm{vs.} \frac{\partial}{\partial u^{1}}$) they map to 1. However, by means of $x^{1} = u^{1}$, I would expect their differentials or exterior derivatives, respectively, to be equal as well, i.e., $\mathrm{d}x^{1} = \mathrm{d}u^{1}$. Does anyone know where there is the obvious fallacy here?
The way in which you used $x^1$ originally is as a function on $M,$ i.e. $x^1: M \to\mathbb{R}.$ This can be seen in the definition,
$$\varphi(u^{1}) = x^{1}(u^{1})\,e_{1} + x^{2}(u^{1})\,e_{2}.$$
The sense that $x^1 = u^1$ is in the sense that they agree as functions on $M.$ Specifically let $u^1: M \to \mathbb{R}$ be the coordinate function and observe that for all $p \in M,$
$$x^1(p) = u^1(p).$$
In this strict sense,
$$d x^1 = d u^1,$$
where $d u^1$ is the smooth coordinate one-form. Recognize that $x^1$ is strictly viewed as a function on $M$ and not on $\mathbb{R}^2$, so there is no contradiction here in terms of the vectors they take: both $dx^1$ and $du^1$ operate on vectors in $TM.$ The confusion may be arising from the fact that we have another coordinate function $\bar x^1: \mathbb{R}^2 \to \mathbb{R}$ on the ambient manifold $\mathbb{R}^2$ which agrees with the preceding $x^1$ on the manifold $M$. That is given the insertion map $\iota: M \to \mathbb{R}^2$,
$$\iota^* \bar x^1 = x^1.$$
The only correspondence they have is on the manifold $M$ where we of course have,
$$d( \iota^* \bar x^1 ) = \iota^* d \bar x^1 = d x^1.$$
It is $d\bar x^1$ that takes in vectors in $T\mathbb{R}^2$, while the pullback $\iota^* d\bar x^1 = dx^1$ that takes in vectors of $T M$. Of course, you could just as easily abuse notation and let $d\bar x^1 = d x^1$ but then you have to worry about these issues when they arise.