On page 20 of Bott and Tu's "Differential Forms in Algebraic Topology", they say
Let $x_1,\ldots,x_n$ be the standard coordinate system (on $\mathbb{R}^n$) and let $u_1,\ldots,u_n$ be a new coordinate system on $\mathbb{R}^n,$ i.e. there is a diffeomorphism $f:\mathbb{R}^n\to\mathbb{R}^n$ such that $u_i=x_i\circ f.$
In the next line, they let $g:\mathbb{R}^n\to \mathbb{R}$ be smooth and use the symbol
$$\frac{\partial g}{\partial u_i}.$$
Question. What does this mean?
Some thoughts. We have a (non-canonical) $\mathbb{R}$-linear isomorphism $\mathbb{R}^n \to \mathcal{L}(\mathbb{R}^n,\mathbb{R})$ given by
$$v \mapsto \langle \cdot , v \rangle$$
where $\langle \;, \,\rangle$ is the standard inner product on $\mathbb{R}^n.$ This is the sense in which the standard coordinate maps $\{x_i\}$ correspond to the standard basis $\{e_i\}$ of $\mathbb{R}^n.$ So given any $\alpha \in \mathcal{L}(\mathbb{R}^n,\mathbb{R}),$ I can define
$$\frac{\partial g}{\partial \alpha}$$
to mean the directional derivative of $g$ in the direction of the vector corresponding to $\alpha$ under the above isomorphism.
However, it's not true in general that $u_i$ is $\mathbb{R}$-linear and so I can't interpret in this naïve way...
In which case, do we just define
$$\frac{\partial g}{\partial u_i} = \frac{\partial (g\circ f^{-1}) }{\partial x_i}\circ f?$$
The vector fields $\frac{\partial}{\partial x_i}$ are a basis of the tangent space to $\Bbb{R}^n$, and represent directional derivative operators. The change-of-coordinate map $f \colon \Bbb{R}^n \to \Bbb{R}^n$ has a differential $df$, which is a linear map $T\Bbb{R}^n \to T\Bbb{R}^n$ from the tangent space of $\Bbb{R}^n$ (in the original $x$-coordinates) to itself (but now in the $u$-coordinates), represented in these coordinates by the Jacobian matrix $\displaystyle \left( \frac{\partial u_i}{\partial x_j} \right)_{i,j=1}^n$. We can also form the inverse of this matrix, $\displaystyle df^{-1} = \left( \frac{\partial x_i}{\partial u_j} \right)$. You can use this matrix to identify $\frac{\partial}{\partial u_i}$ with its image under $df^{-1}$. So the operators $\frac{\partial}{\partial u_i}$ can be thought of, in the original $x$-coordinates, as directional derivatives, given by the columns of this matrix $df^{-1}$.
A good example to keep in mind is polar coordinates. Think of $u_1,u_2 = r,\theta$. Then in your notation, $f^{-1}(r,\theta) = (r\cos(\theta),r\sin(\theta))$. The Jacobian matrix is $$ df^{-1} = \left( \begin{array}{cc} \cos(\theta) & -r\sin(\theta) \\ \sin(\theta) & r\cos(\theta) \end{array} \right) $$
So $\frac{\partial}{\partial r}$ can be thought of as the vector $(\cos(\theta), \sin(\theta))$, the unit vector pointing radially outward (this is the first column of $df^{-1}$). Similarly, $\frac{\partial}{\partial \theta}$ can be thought of as the second column, which is $r(-\sin(\theta),\cos(\theta))$. This is tangent to the circle through the given point, and its length is proportional to $r$, the distance from the origin.