Question about change-of-basis matrix I read a chapter in my algebra book, about change-of-basis matrix.

Let $V$ a vectorial space over a field $\mathbb{K}$ and $B=\{e_1,e_2,...,e_n\}, B'=\{e_1',e_2',...,e_n'\}$ two basis of this vectorial space. Then, every element of basis $B'$ can be written as linear combination of vectors from $B$.

$e_1' = u_{11}e_1+u_{21}e_2+\dots+u_{n1}e_n$ 
$e_2' = u_{12}e_1+u_{22}e_2+\dots+u_{n2}e_n$ 
$\dots \dots\dots\dots\dots\dots\dots\dots$ 
$e_n' = u_{1n}e_1+u_{2n}e_2+\dots+u_{nn}e_n$ 
Thus, $U=\begin{pmatrix}
u_{11} && u_{12} && \dots && u_{1n} \\
u_{21} && u_{22} && \dots && u_{2n} \\
\vdots && \vdots && \ddots  && \vdots \\
u_{n1} && u_{n2} && \dots && u_{nn}
\end{pmatrix}$ is the change of basis matrix from $B$ to $B'$.

My question is:
Why this matrix is named " from $B$ to $B'$ " if we have relation: $[v]_{B} = U * [v]_{B'}$? Shouldn't it be called "from $B'$ to $B$?
Sorry for eventually typos. I've translated this fragment from another language to english.
 A: 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...
A: 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.
