Question about change-of-basis matrix

The author is correct.... In order to see this let's apply the change of basis matrix $U$ to an arbitrary basis element in $B$;

$$U\circ\ e_1 = \left[ {\begin{array}{cccc} u_{11} & u_{12} & \cdots & u_{1n}\\ u_{21} & u_{22} & \cdots & u_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ u_{n1} & u_{n2} & \cdots & u_{nn}\\ \end{array} } \right] \circ \begin{bmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{bmatrix} = \begin{bmatrix} u_{11} \\ u_{21} \\ \vdots \\ u_{n1} \end{bmatrix} $$

$$=\space u_{11}\begin{bmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{bmatrix}+u_{21}\begin{bmatrix} 0 \\ 1 \\ \vdots \\ 0 \end{bmatrix}+\dots u_{n1}\begin{bmatrix} 0 \\ 0 \\ \vdots \\ 1 \end{bmatrix}=u_{11}e_1+u_{21}e_2+ \dots u_{n1}e_n= e'_1$$

$$ \therefore U\circ\ e_{1} = e'_{1}$$

So the matrix $U$ does, indeed, transform basis vectors from $B$ to $B'$ as the author stated.

$\textbf{EDIT:}$

Per your example, let's determine the change of basis matrix U from the basis $B_1$ to the basis $B_2$.

$$e'_1=U_{B_1 \rightarrow B_2}\circ e_1=\left[ {\begin{array}{cc} u_{11} & u_{12}\\ u_{21} & u_{22} \end{array} } \right]\circ\begin{bmatrix} 2 \\ 3 \\ \end{bmatrix} =\begin{bmatrix} {2u_{11}+3u_{12}}\\ {2u_{21}+3u_{22}} \\ \end{bmatrix}=\begin{bmatrix} 1\\ 4\\ \end{bmatrix}$$

$$e'_2=U_{B_1 \rightarrow B_2}\circ e_2=\left[ {\begin{array}{cc} u_{11} & u_{12}\\ u_{21} & u_{22} \end{array} } \right]\circ\begin{bmatrix} 8 \\ 5 \\ \end{bmatrix}=\begin{bmatrix} {8u_{11}+5u_{12}} \\ {8u_{21}+5u_{22}} \\ \end{bmatrix}=\begin{bmatrix} 3 \\ 7 \\ \end{bmatrix}$$

So we now have 2 sets of 2 equations with 2 unknowns each which are completely determined. This system gives us the following which you are free to check...

$$U_{B_1 \rightarrow B_2}= {1\over7}\left[ {\begin{array}{cc} {2} & {1}\\ {1\over2} & {9} \end{array} } \right]$$

$\textbf{EDIT 2:}$

Let's consider a scenario where I am using a certain 2D-basis $B$, while my friend is using a basis which, from my perspective, looks like $B'=\{\begin{bmatrix} 2 \\ 3 \\ \end{bmatrix},\begin{bmatrix} 8 \\ 5 \\ \end{bmatrix} \}$

Let's say my friend refers to the vector $\begin{bmatrix} 1 \\ 0 \\ \end{bmatrix}$ in his coordinate system. In order for me to understand what vector my friend is referring to in my coordinate system I have to compose it with the matrix whose columns are his basis vectors.

$$U\circ \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix}=\left[ {\begin{array}{cc} 2 & 8\\ 3 & 5 \end{array} } \right]\circ \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} = 1\cdot \begin{bmatrix} 2 \\ 3 \\ \end{bmatrix}+0\cdot \begin{bmatrix} 8 \\ 5 \\ \end{bmatrix}$$

So the vector which my friend "sees" as $\begin{bmatrix} 1 \\ 0 \\ \end{bmatrix}$ looks like $\begin{bmatrix} 2 \\ 3 \\ \end{bmatrix}$ in my basis $B$ which is clearly...

$$e'_{1}=\begin{bmatrix} u_{11} \\ u_{21} \\ \end{bmatrix} = u_{11}e_1+u_{21}e_2$$

As required...

One can view this process in two different ways:

$1).\ $ using the definition of a change of basis matrix, one has that $U,$ viewed as a linear isomorphism of $V$, satisfies $Ue_i=e'_i.$ That is all the author is saying here.

The confusion arises because

$2).\ U,$ the matrix, also transforms the coordinates of a fixed vector $v$ in the basis $B'$ to the coordinates of that same vector in the basis $B.$ To see why, refer to your own calculation:

$e_1' = u_{11}e_1+u_{21}e_2+\dots+u_{n1}e_n$
$e_2' = u_{12}e_1+u_{22}e_2+\dots+u_{n2}e_n$
$\dots \dots\dots\dots\dots\dots\dots\dots$
$e_n' = u_{1n}e_1+u_{2n}e_2+\dots+u_{nn}e_n$

and take a fixed vector $v$ and express it as

$\tag1 v=a_1e_1'+\cdots + a_n e_n'.$

The coordinates of $v$ in the basis B' are $[a_1,\cdots, a_n].$ Now, substitute for the $e_i'$ into $(1)$ and you will obtain the coordinates of $v$ in the $B$ basis. To finish, observe how $U$ encodes the change of coordinates. Note that the vector $v$ is fixed. It's its coordinate expression that changes, as the basis changes.