Sum of two subspaces: representing it with equations I found the following excercise:

Let $W_1 = \{(x_1, ..., x_6) : x_1 + x_2 + x_3 = 0, x_4 + x_5 + x_6 = 0 \}$. Let $W_2$ be the span of $S := \{(1, -1, 1, -1, 1, -1), (1, 0, 2, 1, 0, 0), (1, 0, -1, -1, 0, 1),  (2, 1, 0, 0, 0, 0)\}$.
Give a base, a dimension and an equation representation of $W_1 + W_2$

I'm new to the concept of sum of subspaces. But as I understand it, the first step would be to note that any $\textbf{x} = (x_1, ..., x_6) \in W_1$ satisfies
\begin{equation*}
    \begin{cases}
        x_1 = -x_2 - x_3 \\
        x_4 = -x_5 - x_6
    \end{cases}
\end{equation*}
so that its general form is
\begin{equation*} \textbf{x} =
    \begin{pmatrix}
        x_1 \\
        x_2 \\
        x_3 \\
        x_4 \\
        x_5 \\
        x_6
    \end{pmatrix} = \begin{pmatrix}
         -x_2 - x_3 \\
        x_2 \\
        x_3 \\
        -x_5 - x_6 \\
        x_5 \\
        x_6
    \end{pmatrix}
\end{equation*}
We also know any $\textbf{y} \in W_1$ is of the general form
\begin{align*} \textbf{y} =
    \begin{pmatrix}
         y_1 \\
        y_2 \\
        y_3 \\
        y_4 \\
        y_5 \\
        y_6
    \end{pmatrix} = \begin{pmatrix}x + y +z + 2w \\
    -x + w \\ 
    x + 2y - z \\
    -x + y - z \\
    x\\
    -x + z\end{pmatrix}
\end{align*}
Then for generals $\mathbf{x}, \mathbf{y}$ we have
\begin{align*}
    \textbf{x} + \textbf{y} &= \begin{pmatrix}
        x + y +z + 2w + (-x_2 - x_3)\\
    -x + w + x_2\\ 
    x + 2y - z + x_3\\
    -x + y - z + (-x_5 - x_6)\\
    x + x_5\\
    -x + z + x_6
    \end{pmatrix} 
\end{align*}
One can then conclude
$$W_1 + W_2 = \Big\{\big(x + y + z + 2w - x_2 - x_3\big), \big(-x + w + x_2 \big), \big(x +2y - z + x_3 \big), \big(-x + y - z - x_5 - x_6 \big), \big(x + x_5 \big), \big(-x + z + x_6 \big) \mid x, y, z, x_2, x_3 \in \mathbb{R}  \Big\}$$
But what would be an representation via equations of this system? I'm very new to linear algebra so go easy on me!

You already found that a vector $(a,b,c,d,e,f)$ belongs to $W_1+W_2$ if and only if there exist real numbers $x_2,x_3,x_5,x_6,x,y,z,w$ such that
$$\begin{cases}a&=x+y+z+2w-x_2-x_3\\b&=-x+w-x_2\\c&=x + 2y - z + x_3\\
    d&=-x + y - z-x_5 - x_6\\
    e&=x + x_5\\
    f&=-x + z + x_6.
\end{cases}$$
Using for instance the 2nd, 3rd, 5th, and 6th equations to first eliminate the $x_i$'s, this is equivalent to the existence of real numbers $x,y,z,w$ such that
$$\begin{cases}a&=3x+3y+w+b-c\\
    d&=-x + y-e -f
\end{cases}$$
For any $(a,b,c,d,e,f)\in\Bbb R^6,$ such numbers $x,y,z,w$ always exist (e.g. choose $x,z$ arbitrarily and take $y=d+e+f+x,$ and $w=a-b+c-3x-3(d+e+f+x)$) hence
$$W_1+W_2=\Bbb R^6.$$
You could see it more rapidly since you already knew that $\dim(W_1\cap W_2)=2$: $\dim(W_1)$ is $6-2=4$ (since its 2 equations are clearly independent), $\dim(W_2)=4$ (since its 4 generating vectors are - less clearly - independent) hence by Grassmann's formula,
$$\dim(W_1+W_2)=4+4-2=6.$$
