Here I assume you are talking about the transport equation on $\Omega \subset \mathbb{R}^n \times \mathbb{R}$, so that $u: \Omega \longrightarrow \mathbb{R}.$ The transport equation can be rewritten as $(\vec{b},1) \cdot D_{(x,t)} u = 0,$ which is the directional derivative of $u$ in the direction of $(\vec{b},1)$. So the transport equation tells us that $u$ is constant along lines parallel to $(\vec{b}, 1)$ in $\mathbb{R}^{n+1}$.
As stated, however, the transport equation is not well-posed, as any constant function is a solution. One way to fix this is to introduce an initial condition: $\begin{cases} u_t + \vec{b} \cdot D_x u = 0, \\ u(x,0) = g(x) \end{cases}$ for some $g: \mathbb{R}^n \longrightarrow \mathbb{R}$. Now given $(x,t) \in \Omega)$, let $(x_0,0)$ be the point in $\mathbb{R}^n \times \{0\}$ on the line through $(x,t)$ parallel to $(\vec{b},t)$. Since $u$ is constant on such lines, we have that $u(x,t) = u(x_0,0) = g(x_0) = g(x - t\vec{b}).$