Matrix of the perpendicular Projection I am bit doubtful about my reasoning so I hope there's someone that will correct me if I'm wrong on the steps 1 to 3.
Given is: 
$W = span(v_1, v_2)$ 
with: 
$v_1 = \begin{bmatrix} 1\\0\\1\\0  \end{bmatrix}, v_2 = \begin{bmatrix} 1\\-1\\1\\-1  \end{bmatrix}$
Steps:
i)
I use the Gram-Schmidt process to find an orthonormal  basis which, using $v_1$ as $b_1$(first vector), is 
$W = span\{1/\sqrt{2}\begin{bmatrix}1\\0\\1\\0 \end{bmatrix};1/\sqrt{2}\begin{bmatrix}0\\-1\\0\\-1\end{bmatrix}  \}$
ii)
I find the the Matrix of the orthogonal reflection relative to the canonical basis:
$M = 1 - 2n^Tn$,  I could choose either of the two vector, and I choose $b_1$.
$M = I - 2(1/\sqrt{2})(1/\sqrt{2})\begin{bmatrix}1\\0\\1\\0 \end{bmatrix}\begin{bmatrix}1&0&1&0\end{bmatrix}$
$M = \begin{bmatrix}0&0&1&0\\0&1&0&0\\1&0&0&0\\0&0&0&1\end{bmatrix}$
iii)
I should compute the Matrix of the perpendicular projection relative to the canonical basis, $proj_w : V \to V$ on $W$.
I thought I should compute $<x,b_1>b_1 + <x,b_2>b_2$ with $b_j$ being a vector of the orthonormal basis found at step 1 and $x$ being $x =(x_1,x_2,x_3,x_4)$  and that should be it, but I'm very unsure. 

Yes, your reasoning is correct for iii). The columns of the matrix are the images of the canonical basis, so the first column is
$$\langle e_1, b_1 \rangle b_1 + \langle e_1, b_2 \rangle b_2 = \frac{1}{\sqrt{2}} b_1$$
and similarly for $e_2, e_3, e_4$.
