Finding the basis transition matrix $A$ from $U$ into $V$ satisfying $V=AU$.

Consider the vector space $\mathbb{R^3}$ and the bases $U=\{e_1,e_2,e_3\}$ and $V=\{e_1, e_1+e_2, e_1+e_2+e_3\}$. I want to find the basis transition matrix $A$ from $U$ into $V$ satisfying $V=AU$. Also i want to find the coordinate vectors of the given vector $(1,2,3) \in \mathbb{R^3}$ with respect to the bases $U$ and $V$, respectively, and relate these vectors with the matrix $A$.

What I have done: I define a map $T:U \longrightarrow V$ such that $T(e_1)=e_1$, $T(e_2)=e_1 + e_2$ and $T(e_3)=e_1 + e_2 + e_3$. I've denoted $e_1=v_1$, $e_1+e_2=v_2$ and $e_1 + e_2 + e_3 =v_3$. Then, I've obtained that $e_1=v_1$, $e_2=v_2-v_1$ and $e_3=v_3-v_2$. I guess that the transition matrix A will be that $A= \left[ {\begin{array}{cc} 1 & 1 & 1\\ 0 & 1 & 1\\ 0 & 0 & 1\\ \end{array} } \right]$. Here I am confused. I suppose, something is wrong. Is there any algorithm to solve this type questions?

The way I always found convenient to find the matrix of basis transition is to consider it as the identity map $$\text{Id}: \mathbb{R}^3 \to \mathbb{R}^3,$$ where the first space has basis $U$ and the second space has basis $V$. If you then form the matrix as you usually do, you should find the matrix $$\begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1\\ 0 & 0 & 1\end{pmatrix}$$ as you did. How did we find this? Well, if you form the matrix corresponding to some linear transformation, you consider the image of the basisvectors in $U$ under that transformation and look for the coordinates of these images with respect to the basis $V$. So the coordinates of the image of the first basisvector of $U$ form the first column, the coordinates of the image of the second basisvector of $U$ form the second column, and so on.
I am not quite sure what you did with the part where you considered $T$, nor how you could conclude from it how $A$ looks.