How does a vector b in the column space come from a vector in the row space? I'm working through Gilbert Strang's Introduction to Linear Algebra book and am really confused by a paragraph from chapter 4.1 titled 'Orthogonality of the Four Subspaces'. The paragraph is as follows:

Every vector goes to the column space! Multiplying by A cannot do anything else. More than that: Every vector $b$ in the column space comes from one and only one vector $x_r$ in the row space. Proof: If $Ax_r = Ax'_r$, the difference $x_r - x'_r$ is in the nullspace. It is also in the row space, where $x_r$ and $x'_r$ came from. This difference must be the zero vector, because the nullspace and row space are perpendicular. Therefore $x_r = x'_r$.

Further on in the book an exercise is given, where we have to demonstrate this using the following figure: Two pairs of orthogonal subspaces, with the following matrix: $A = \begin{bmatrix}1 & 2\\3 & 6\end{bmatrix}$. The column space of the matrix is: $(1, 3)$, and its row space is: $(1, 2)$. If I multiply A with the randomly chosen $x$ vector: $(1, 1)$, I arrive at $b = (3, 9)$. However, this $b$ seems unable to be recreated using a multiple of the row space vector: $(1, 2)$. I'm really confused by this. I also feel like I'm missing the meaning of the proof and am not familiar with the $'$ symbol in $Ax'_r$. Does it mean the transpose?
Any help would be greatly appreciated! 

The $'$ doesn't denote transpose in this case; it's just an adornment to indicate that $x_r$ and $x_r'$ are two vectors—the author have equally well called them $x_r$ and $y_r$.
Note that $A\begin{bmatrix}1\\2\end{bmatrix} = \begin{bmatrix}5\\15\end{bmatrix}$, which is a multiple of $b=\begin{bmatrix}3\\9\end{bmatrix}$. Since matrix multiplication is linear, we can set $x_r=\frac35\begin{bmatrix}1\\2\end{bmatrix} = \begin{bmatrix}3/5\\6/5\end{bmatrix}$, which is in the row space, and for which $Ax_r=\frac35\begin{bmatrix}5\\15\end{bmatrix} = b$.
