Given a linear map $T \colon V \rightarrow V$ and a choice of ordered basis $\mathcal{B} = (v_1, \ldots, v_n)$ for $V$, the matrix $[T]_{\mathcal{B}}$ representing $T$ using the basis $\mathcal{B}$ for the domain and range is by definition the matrix whose columns are the vectors $[Tv_1]_{\mathcal{B}}, \ldots, [Tv_n]_{\mathcal{B}}$.
Since $T$ is a projection to the $x$ axis in parallel to $y = 2x$, it is natural to take a basis for both subspaces and see how $T$ acts on in. As you wrote, we can take $\mathcal{C} = (v_1, v_2)$ where
$$ v_1 = \left( \begin{matrix} 1 \\ 0 \end{matrix} \right), v_2 = \left( \begin{matrix} 1 \\ 2 \end{matrix} \right) $$
to be our basis. With respect to this basis, since $T$ is a projection, it acts as $T(a_1v_1 + a_2v_2) = a_1v_1$. That is, if you decompose a vector $v$ as $v = a_1v_1 + a_2v_2$ the projection just forgets the component of $v$ in the direction of $v_2$. With respect to this basis, we have
$$ [T(v_1)]_{\mathcal{C}} = [v_1]_{\mathcal{C}} = \left( \begin{matrix} 1 \\ 0 \end{matrix} \right), [T(v_2)]_{\mathcal{C}} = [0]_{\mathcal{C}} = \left( \begin{matrix} 0 \\ 0 \end{matrix} \right) $$
so
$$ [T]_{\mathcal{C}} = \left( \begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix} \right). $$
However, you are asked to find $[T]_{\mathcal{B}}$ where $\mathcal{B} = (e_1, e_2)$ where $e_i$ are the standard basis vectors. Thus, we need to find $[Te_1]_{\mathcal{B}}$ and $[Te_2]_{\mathcal{B}}$. To find $T(e_1)$ is easy since $e_1$ lies on the $x$ axis and $T$ projects to the $x$ axis and so leaves $e_1$ invariant. Thus, $T(e_1) = e_1$ and
$$ [T(e_1)]_{\mathcal{B}} = [e_1]_{\mathcal{B}} = \left( \begin{matrix} 1 \\ 0 \end{matrix} \right). $$
However, in order to understand what is $T(e_2)$, we have to decompose $e_2$ as $e_2 = a_1v_1 + a_2v_2$ and then $T(e_2) = a_1v_1$. Indeed, if
$$ e_2 = \left( \begin{matrix} 0 \\ 1 \end{matrix} \right) = a_1v_1 + a_2v_2 = a_1 \left( \begin{matrix} 1 \\ 0 \end{matrix} \right) + a_2 \left( \begin{matrix} 1 \\ 2 \end{matrix} \right) = \left( \begin{matrix} a_1 + a_2 \\ 2a_2 \end{matrix} \right) $$
then by comparing coefficients we obtain the system of linear equations
$a_1 + a_2 = 0, 2a_2 = 1$ whose unique solution is $a_2 = \frac{1}{2}$, $a_1 = -\frac{1}{2}$.
Thus,
$$ [T(e_2)]_{\mathcal{B}} = \left[ -\frac{1}{2} v_1 \right]_{\mathcal{B}} = \left[ \left( \begin{matrix} -\frac{1}{2} \\ 0 \end{matrix} \right) \right]_{\mathcal{B}} = \left( \begin{matrix} -\frac{1}{2} \\ 0 \end{matrix} \right)$$
and
$$[T]_{\mathcal{B}} = \left( \begin{matrix} 1 & -\frac{1}{2} \\ 0 & 0 \end{matrix} \right). $$
Finally, note that $T$ is not given by
$$ T\left( \begin{matrix} x \\ y \end{matrix} \right) = \left( \begin{matrix} x \\ 0 \end{matrix} \right) $$
(this is the projection to the $x$ axis in parallel to the $y$ axis) but
$$ T \left( \begin{matrix} x \\ y \end{matrix} \right) = [T]_{\mathcal{B}} \left( \begin{matrix} x \\ y \end{matrix} \right) = \left( \begin{matrix} 1 & -\frac{1}{2} \\ 0 & 0 \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) = \left( \begin{matrix} x - \frac{1}{2}y \\ 0 \end{matrix} \right). $$
Best Answer
a) The projection of $v$ on the vector $w$ is $\displaystyle\frac{v.w}{\lVert w\rVert^2}w$. So, the projection of $(1,0)$ is $\displaystyle\left(\frac15,-\frac25\right)$ and the projection of $(0,1)$ is $\displaystyle\left(-\frac25,\frac45\right)$. So, the matrix is$$\begin{bmatrix}\frac15&-\frac25\\-\frac25&\frac45\end{bmatrix}.$$
b) Note that $(2,3)=3(1,1)-(1,0)$. Therefore, $T(2,3)=3T(1,1)-T(1,0)$.