$y + z+ w = 0$ implies that $w = -y - z$. Therefore, vectors in $Y_1$ are of the form $$(x,y,z,-y-z)$$
for some $x,y,z \in \mathbb{R}$.
We have:
$$Y_1 \ni (x,y,z,-y-z) = x(1,0,0,0) + y(0,1,0,-1) + z(0,0,1,-1)$$
so the set $\{(1,0,0,0), (0,1,0,-1), (0,0,1,-1)\}$ spans $Y_1$. Furthermore, it is linearly independent, so it is a basis for $Y_1$.
Similarly, $x + y = 0$ implies $y = -x$ so vectors in $Y_2$ are of the form
$$(x,-x,2w,w)$$
for some $x,w \in \mathbb{R}$.
We have:
$$Y_2 \ni (x,-x,2w,w) = x(1,-1,0,0) + w(0,0,2,1)$$
so the set $\{(1,-1,0,0), (0,0,2,1)\}$ spans $Y_2$. Furthermore, it is linearly independent so it is a basis for $Y_2$.
So, the general method would be to use the conditions to establish how a vector in your subset look like. It will be of the form
$$(\alpha_1x + \beta_1y + \gamma_1z + \delta_1w, \alpha_2x + \beta_2y + \gamma_2z + \delta_2w, \alpha_3x + \beta_3y + \gamma_3z + \delta_3w, \alpha_4x + \beta_4y + \gamma_4z + \delta_4w)$$
and write it as a linear combination of some constant set of vectors. That set of vectors spans your subspace. Now reduce it to a linearly independent set. It will then be a basis.
The rows of $A$ are $x$ and $y$, and so $S$ is the row space of A:$$S=\{\alpha x+\beta y:\ \alpha,\beta\in\mathbb{R}\}$$
Therefore, $S^⊥=\{z\in\mathbb{R^3:\ (\alpha x+\beta y+z)=0}\}$ for all $\alpha,\beta$ ; in particular,
$(x, z) = 0$ and $(y, z) = 0$. But the definition of matrix multiplication says that$$Az=\begin{bmatrix} (x,z) \\ (y,z)\end{bmatrix}=0$$Hence, $z\in N(A)$
For the reverse inclusion, suppose that $z \in N(A)$. Now $Az = 0$. As above, the definition of matrix multiplication gives $(x, z) = 0 = (y, z)$. Hence, for all
scalars $\alpha,\beta$ we have$$(\alpha x+\beta y,z)=\alpha(x,z)+\beta(y,z)=0$$ So, $z\in S^⊥$
Best Answer
From the answer: since $y_1=y_2-y_3$, one has that the solution may be written as $$ \left( \begin{array}{c} y_1\\y_2\\y_3 \end{array}\right)= \left( \begin{array}{ccc} y_2-y_3\\y_2\\y_3 \end{array}\right)= \left( \begin{array}{ccc} y_2 \\ y_2 \\ 0 \end{array}\right)+\left( \begin{array}{ccc} -y_3 \\ 0 \\ y_3 \end{array}\right) = y_2\left(\begin{array}{c}1 \\ 1 \\ 0 \end{array}\right)+y_3\left( \begin{array}{ccc} -1 \\ 0 \\ 1 \end{array}\right) $$ that is, it is a linear combination of the last two column vectors. Since it's a linear combination, it need scalars, which will be $y_2$ and $y_3$. Now, just give those scalars names: $y_2=\alpha$ and $y_3=\beta$.
Note, however that, since you wrote $(y_2-y_3,y_1+y_3,y_2-y_1)^T$, it is not wrong, but you will not be able to find a basis for the vector space writing it in this form (you'll end up finding three vectors, but in true you need only two). Best way is to let one of the scalars (in this case, $y_1$) be written as dependent of the last two (in this case, $y_2$ and $y_3$) and separate them into a linear combination as I did above.