The linear combinations of the vectors $v,w$ can be denoted $span(v,w)=kv+cw$, where $v,w \in V$ and $k,c \in \mathbb{F}$ (scalars).
If the two vectors are linearly independent, informally consider the follow scenario:
Let $V = \mathbb{R^2}$. Let two vectors $v,w \in V$. Set $v = (x,0)$ and $w = (0,y)$. There is no scalar $k \in \mathbb{F}$, that when multiplied by the vector $v = (x,0)$, yields a vector of the form $w = (0,y)$. The two independent vectors, when combined linearly as $kv+cv$, make up a whole plane or $\mathbb{R^2}$. You can think of these vectors as the $x-axis$ and $y-axis$ coming together to form the $xy-plane$. When taking the set of all linear combinations of the vectors $v$ and $w$ we can obtain any other vector of the form $(x,y)$, and thus $span(v,w) = \mathbb{R^2}$.
Formally, the only combination $kv+cw$ that yields the $0$ vector is by setting each $k,v \in \mathbb{F}$ equal to $0$. Is is clear that the only scalars that satisfy the linear combination $k(x,0) + c(0,y) = (0,0)$ for the $0$ vector are $k = c = 0$, which implies that the two vectors in $\mathbb{R^2}$ are independent and span the plane.
Now, consider a separate case, where the vectors $v,w \in V$ are $(x,0)$ and $(2y,0)$, respectively. Let $span(v,w) = a(x,0) + b(2y,0)$ which denotes all possible linear combinations of the vectors. However, it is apparent that the $0$ can be obtained by more than just the "obvious" way which means setting $a = b = 0$ to obtain $(0,0) + (0,0) = (0,0)$. Setting $y = \frac{-1}{2}x$, and we get $a(x,0)+b(-x,0)$ which equals $0$ for $a = b = 0$ or $a = b = 1$, which shows that the two vectors are dependent and as a result do not span $\mathbb{R^2}$, so their linear combinations do not fill a plane.
It's not that $v$ and $w$ can't lie on the same line. It's just that if we have two vectors are on the same line, then they are two linearly dependent vectors in $\mathbb{R^2}$, and as a result, do not fill a plane.
1-2) This construction works because the columns are on the left and the rows are on the right. I imagine it was conceived by thinking about how matrix multiplication on the left and right produces a vector. On a more subtle note, vectors are only matrices when you are working in some basis. More generally, vectors do not a priori have a default matrix representation.
3) I you multiply $$B_{m \times n} = \begin{bmatrix}
\vert & & \vert \\
\mathbf{c_1} & \cdots & \mathbf{c_s} \\
\vert & & \vert \\
\end{bmatrix}_{m \times s}
\begin{bmatrix}
-- & \mathbf{{r_1}^T} & -- \\
& ... & \\
-- & \mathbf{{r_s}^T} & -- \\
\end{bmatrix}_{s \times n}
= \sum\limits_{1 \le i \le s}\mathbf{c_s{r_s}^T} $$ by a column vector on the right, the column times $\begin{bmatrix}
-- & \mathbf{{r_1}^T} & -- \\
& ... & \\
-- & \mathbf{{r_s}^T} & -- \\
\end{bmatrix}_{s \times n}$ gives a new column. Multiplying this column by $\begin{bmatrix}
\vert & & \vert \\
\mathbf{c_1} & \cdots & \mathbf{c_s} \\
\vert & & \vert \\
\end{bmatrix}_{m \times s}$ gives you an element of the column space. If you multiply a row vector on the left of $B_{m\times n}$ it similarly gives you an element of the row space.
Best Answer
For the first question, perhaps you could elaborate on which part of the proof you do not understand?
For the second one, here's a simple counter-example: let $A$ be some non-zero matrix, and let $B = -A$. Then $$\mathrm{colspace}(A) = \mathrm{colspace}(B),$$and $$\mathrm{colspace}(A) + \mathrm{colspace}(B) = \mathrm{colspace}(A) \not= \{0\},$$but $$\mathrm{colspace}(A+B) = \mathrm{colspace}(\mathrm{zero matrix}) = \{0\}.$$