This is just a (well known) formal property of matrix multiplication. From the definition, the $(i,j)$ entry of the matrix product $AB$ is the same as the unique entry of the product $R_iC_j$ where $R_i$ is row $i$ of $A$ (a $1\times n$ matrix) and $C_j$ is column $j$ of $B$ (a $n\times 1$ matrix). By varying $i$ it follows from this that column $j$ of $AB$ equals $AC_j$, in other words each column of $AB$ can be computed using the corresponding column of $B$ only. This answers you question. Alternatively one can vary $j$ to find that row $i$ of $AB$ equals $R_iB$, so each row of $AB$ can be computed using the corresponding row of $A$ only. Both points of view are occasionally useful.
To begin with, the Vector Space of Matrices $\mathscr{M}_{mn}(\mathbb{R}^4)$ is not abelian under matrix product! It is a fundamental error to change sides of a matrix by "inverting it" in an equation.
And keep in mind: you can see obviously that $\mathbb{R}^2$ can be extended so to be isomorphic to $\mathbb{R}^4$. It is why they indicated that the image was a projection in fact.
Now let's calculate:
The first intuition is to write down what your matrix operations look like; to find out if the product is well-defined somehow: this will eliminate some big issues in your thoughts.
$$ F(\left (
\begin{matrix}
x_1 \\
x_2 \\
x_3 \\
x_4
\end{matrix}
\right )
)=\left (
\begin{matrix}
x_1 \\
x_2
\end{matrix}
\right )
$$
ie.
$$ \left (
\begin{matrix}
a_{11} & a_{12} & a_{13} & a_{14} \\
a_{21} & a_{22} & a_{23} & a_{24}
\end{matrix}
\right )
.\left (
\begin{matrix}
x_1 \\
x_2 \\
x_3 \\
x_4
\end{matrix}
\right )
=\left (
\begin{matrix}
x_1 \\
x_2
\end{matrix}
\right )
$$
when we introduce the Matrix of elements $(a_{ij})$ as the Matrix associated to the linear transformation. We must have a $2x4$ matrix for this transformation.
When you do the calculations, you'll find:
$$ \begin{cases}
a_{11}x_{1} + a_{12}x_{2} + a_{13}x_{3} + a_{14}x_{4} = x_1 \\
a_{21}x_{1} + a_{22}x_{2} + a_{23}x_{3} + a_{24}x_{4} = x_2
\end{cases} $$
ie.
$$ \begin{cases}
(a_{11}-1)x_{1} + a_{12}x_{2} + a_{13}x_{3} + a_{14}x_{4} = 0 \\
a_{21}x_{1} + (a_{22}-1)x_{2} + a_{23}x_{3} + a_{24}x_{4} = 0
\end{cases} $$
But as we're looking for a basis, we know it must be linearly independent (within row vctors). So that, we know, every coefficients are zero, with $(a_{11}-1)=0$ and $(a_{22}-1)=0$ so that we have $a_{11}=1$ and $a_{22}=1$, over row vectors.
Thus, we conclude that one basis for this linear transformation in $\mathbb{R}^4$ is:
$$
\left (
\begin{matrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0
\end{matrix}
\right )$$
Best Answer
If $X_2$ is the second column of $X$ then $X_2 = X\left[ \begin{array}{c} 0 \\ 1 \end{array} \right],$ so $X(X'X)^{-1}X'X_2 = X(X'X)^{-1}X'X\left[ \begin{array}{c} 0 \\ 1 \end{array} \right]$ and you get a cancellation that leaves only $X \left[ \begin{array}{c} 0 \\ 1 \end{array} \right],$ which is $X_2.$