Transport equation solution clarification and alternative

Consider the following problem:

Problem 1. Write down an explicit formula for a function $$u$$ solving the initial-value problem $$\left\{\begin{array}{rcl} u_t+b\cdot Du+cu=0 & \text{on} & \mathbb{R}^n\times(0,\infty),\\ u= g & \text{on} & \mathbb{R}^n\times\{t=0\}. \end{array}\right.$$ Here $$c\in\mathbb{R}$$ and $$b\in\mathbb{R}^n$$ are constants.

Sol: Fix $$x$$ and $$t$$, and consider $$z(s):=u(x+bs,t+s)$$. Then \begin{align*} \dot z & =b\cdot Du+u_t\\ & =-cu(x+bs,t+s)\\ & =-cz(s) \end{align*} Therefore, $$z(s)=De^{-cs}$$, for some constant $$D$$. We can solve for $$D$$ by letting $$s=-t$$. Then, \begin{align*} z(-t) & =u(x-bt,0)\\ & =g(x-bt)\\ & =De^{ct} \end{align*} i.e. $$D=g(x-bt)e^{-ct}$$. Thus, $$u(x+bs,t+s)=g(x-bt)e^{-c(t+s)}$$ and so when $$x=0$$, we get $$u(x,t)=g(x-bt)e^{-ct}.$$

In this solution to an exercise in Evans, Why is $$z$$ chosen as such?

Also, what would the solution look like using Fourier transform? Is there a nice way to do it?

The curves of the form $$\gamma(s)=(x+bs,t+s)$$ are the characteristic curves of this equation. You can find the general theory of characteristic curves for first order PDEs on the Evans, or on other books on PDEs.
$$z$$ is $$u$$ when valued on one of this curves. The motivation for defining the characteristic curves this way (in this case) is that the PDE $$\frac{\partial u}{\partial t}+b\cdot\nabla u+ cu =0$$ is equivalent to say that the directional derivative of $$u$$ in direction of the $$n+1$$-dimensional vector $$(b,1)$$ is equal to $$-cu$$. So for $$u$$ being a solution of the PDE is equivalent that all its restrictions on the flow lines of this vector field solve a certain ODE. The flow lines of this vector field are the characteristic curves.