In the appendix of Lovelock's book "Tensors, Differential Forms and Variational Principles" they give a proof of a theorem fundamental to the notion of a tangent vector on a manifold: Part of the proof involves considering a chart $(U,h)$ defined such that $h(p)=(0,\ldots,0)$ and then choosing a function $f\in C^{\infty}_{p}$ such that $q\in U$ is in the domain of $f$, with $h(q)=(u^{1},\ldots,u^{n})$. They then assert that $$f(q)=f(p)+x^{j}(q)f_{j}(q)$$ where $$f_{j}(q)=\frac{\partial f}{\partial x^{j}}\bigg\vert_{p}\equiv \partial_{j}f(p)$$ To establish this they consider a function $g=f\circ h^{-1}$ defined on an open set of $\mathbb{R}^{n}$ that contains the points $(0,\ldots,0)$ and $(u^{1},\ldots,u^{n})$. From this, I see how it follows from the fundamental theorem of calculus that $$g(u^{1},\ldots,u^{n})-g(0,\ldots,0)=\int_{0}^{1}\frac{d}{dt}\left[g(tu^{1},\ldots,tu^{n})\right]dt$$ However, I'm unsure how it follows that $$\frac{d}{dt}\left[g(tu^{1},\ldots,tu^{n})\right]=u^{j}\frac{\partial g}{\partial u^{j}}$$ Is it just that they define the functions $u^{j}:\left[0,1\right]\rightarrow\mathbb{R}^{n}$ such that $u^{j}(t)=tu^{j}$ where $u^{j}\in\mathbb{R}^{n}$, as I could see how this could work via the chain rule, but otherwise I'm a bit confused?!
In addition, I don't quite understand why they introduce a map $h:U\subset M\rightarrow\mathbb{R}^{n}$ defined by $h(p)=(u^{1},\ldots,u^{n})$ and then consider a set of coordinate functions which they define such that $x^{i}=u^{i}\circ h:U\rightarrow\mathbb{R}$ and then $x^{i}(p)=u^{i}$? I always though that given a chart $(U,\phi)$ one defined the map $\phi :U\rightarrow\mathbb{R}^{n}$ such that $x\equiv\phi (p) =(x^{1}(p),\ldots, x^{n}(p))\in\mathbb{R}^{n}$ and so the coordinate functions were already defined through this without the need to introduce a composition of maps?!
$\newcommand{\Reals}{\mathbf{R}}\newcommand{\dd}{\partial}$Briefly, "yes": $g = f \circ h^{-1}$ is "$f$ in the coordinate system $h$", and $u^{j}$ is the $j$th coordinate of $h$, so the equation you ask about is just the chain rule, assuming the summation convention.
In more detail (and to address your final paragraph), $u^{j}:\Reals^{n} \to \Reals$ denotes the $j$th Cartesian coordinate, and $h:U \subset M \to h(U) \subset \Reals^{n}$ is a chart. The coordinate functions of the chart are, by definition, $$ x^{j} = u^{j} \circ h:U \to \Reals. $$ (Caution: Your book seems to have used $(u^{1}, \dots, u^{n})$ to denote both the coordinates of the specific point $q$ and the Cartesian coordinate functions on $h(U)$.)
To say "$g$ is $f$ in the coordinate system $h$" means precisely that $g = f \circ h^{-1}$, or in terms of coordinates, $$ g(u^{1}, \dots, u^{n}) = f(x^{1}, \dots, x^{n}). $$ The preceding equation is a mild abuse of notation; $f$ is defined on $M$, not on a subset of $\Reals^{n}$. What's meant literally is, if $q \in M$ and $h(q) = (u^{1}, \dots, u^{n})$, then the coordinates of $q$ in the chart $h$ are $(x^{1}, \dots, x^{n})$. The preceding equation should be interpreted as $(g \circ h)(q) = f(q)$, or $g\bigl(h(q)\bigr) = (f \circ h^{-1})\bigl(h(q)\bigr)$.
(Similarly, we could write $(u^{1}, \dots, u^{n}) = h(x^{1}, \dots, x^{n})$; in the chart $h$, $h$ itself is essentially the identity map!)
In the same sense, $$ \frac{\dd g}{\dd u^{j}}(u^{1}, \dots, u^{n}) = \frac{\dd f}{\dd x^{j}}(x^{1}, \dots, x^{n}),\quad\text{i.e.,}\quad \frac{\dd g}{\dd u^{j}} = \frac{\dd f}{\dd x^{j}} \circ h^{-1}. $$ The calculation in question does, therefore, amount to the chain rule.