I am trying to understand how $b$ can be any other number than $0$. Since $a$ must be orthogonal to $x$, then their dot product, $a^Tx$, must be equal to $0$. So if $b$ is any other number than $0$, then wouldn't the dot product not be $0$ and therefore not orthogonal?
I am told that $b$ is the shift from the origin but I do not understand why that is.
Let $U\subset \mathbb{R}^n$ be any hyperplane. The argument I'll use works just fine for arbitrary $n\in \mathbb{N}$, but because it might be easier to visualise, consider $n =3$, hence $U$ is a plane. Such plane $U$ can be expressed by its normal vector, let's call it $a\in \mathbb{R}^3$ just as you considered, and it will be expressed as the set of points that are orthogonal to $a$. However, that is not all there is to it.
If your plane contains the zero vector, $U$ will be a vectorspace and your description works fine, but, if $0\not \in U$, you need to specify which of all the planes whose normal vector is $a\in \mathbb{R}^3$ is $U$.
For that matter, assume $p = (p_1,p_2,p_3) \in U$ is a known point of $U$. If $y\in U$, the vector difference $y-p$ belongs to the plane $U_0 = \lbrace \langle a, x\rangle = 0, x\in \mathbb{R}^3\rbrace$, that is, the parallel vector to $U$ that contains the zero vector, i.e., you've got that the points of $y\in U$ satisfy $\langle a, y-p\rangle = 0$ and such thing is $$0 = \langle a, y-p\rangle = (a_1,a_2,a_3)\cdot (y_1-p_1, y_2-p_2, y_3-p_3)^T = \sum_{i=1}^3a_i (y_i-p_i) \\ = \sum_{i=1}^3a_i y_i -\sum_{i=1}^3a_i p_i = \langle a, y\rangle - \langle a,p\rangle$$ where $\langle a,p\rangle\in \mathbb{R}$, so you might just call it $b=\langle a,p\rangle$ and hence your question follows. The answer could've been better written, I guess, but I hope it helps.