If we have
$$u_t + au_x = b$$ with initial condition $u(x, 0) = g(x)$ Let $x$ be a function $t$. Then $$\frac{du}{dt}=\frac{∂u}{∂t}+\frac{∂u}{∂x}\frac{dx}{dt}$$
This is the step that I cannot follow. I can set $x=f(t)$ which gives
$$\frac{du}{dx}=\frac{du}{dt}$$ but can't progess from there.
Your partial differential equaiton is given as:
$$1\cdot u_t + a \cdot u_x = b$$
Your system of characteristic equations is given by:
$$\dfrac {dt}{1}=\dfrac{dx}{a}=\dfrac{du}{b}$$
Using the first equation you will obtain $at-x = c_1$. Using the second equation will give you $bx+ac_2=au \implies u = \frac{b}{a}x+c_2=\frac{b}{a}x+c_2(c_1)=\frac{b}{a}x+c_2(at-x)$
$$\implies u=\frac abx+F(at-x).$$
In the middle, I used that $c_2$ is a function of $c_1=at-x$, this function is then represented by $F$.
Your final goal is to determine the function $F$ such that $u$ is fulfilling $u(x,0)=g(x)$.
Can you complete it from here?