You certainly know that numbers could refer to different kind of object. $5$ could be an area of a triangle, could be the number of apples, could be the value in dollar of an object.
The same works for vectors. Vectors are often used to represent force. But you could use a vector to represent a direction, speed or acceleration, or even they can be used to represent a color in RGB system.
In your picture you have the same vectors, but they have different point of application: the first one is applied in $(0,0)$ the second one, somewhere in the third quadrant.
Vectors could also have 3 dimension, or even n dimension. So don't be confused about them: they are used to represent some concept, but actually they are just "numbers", and they can have the meaning you wish.
I am doing some major guesswork here due to the text being quite confusing. (Where does $w$ come from? Why is $\{rv+sw+x|r,s\in\mathbb R\}$ being classified as a line, when it is clearly a plane?) So here is my attempt to clear your confusion.
Q1. Your solution is incorrect. You need to subtract $\bf x$ from the other vectors.
Q2. Here is also a point I'm confused about. The author is clearly talking about a line in that section of the text. Defining a line as the set $\{kv+x|k\in\mathbb R\}$, we see that we need a direction vector $v$ and a translation vector $x$. The translation vector $x$ defines a starting point of the line, and the direction vector $v$ extends the line in that direction (forwards and backwards). The subtraction $(rv+x)-(sv+x)=(r-s)v$ shows that the direction vector $v$ and the translation vector $x$ are independent, so $x$ can be chosen as any point on the line.
Q3. Subtracting two position vectors gives you the direction vector from one of the points to the other.
Now, to understand the construction of the plane $\{rv+sw+x|r,s\in\mathbb R\}$, we see that there are two direction vectors $v,w$ and a translation vector $x$. The direction vectors $v,w$ span our plane.
To see that $v,w$ are found by subtracting vectors, consider three given (non-collinear) points. They form a triangle. Choose one vertex as the starting point. We calculate the direction vectors from our starting point to the other two vertices. These two vectors are precisely $v,w$.
Finally, the plane $\mathbb R^3$ passing through the points $(1,2,3),(0,2,1),(3,-1,1)$ is given by $$\{r(-1,0,-2)+s(2,-3,-2)+(1,2,3)|r,s\in\mathbb R\}$$
Substituting $(r,s) = (0,0), (1,0),(0,1)$ shows that the plane above does pass through our points.
Best Answer
A line is parametrized by knowing some point on it (given by the vector $\vec x_0$ from the origin) and its direction vector $\vec v$. Then the vector from the origin to an arbitrary point on the line can be written as $$\vec x = \vec x_0 + t\vec v \quad\text{for some value of the scalar } t.$$
Note that if you know two points on the line, this corresponds to such expressions for two different values of $t$, and so you recover the direction vector (or a nonzero scalar multiple of it) by subtracting the two vectors: If $\vec x_1 = \vec x_0 + t\vec v$ and $\vec x_2 = \vec x_0 + s\vec v$ (where $s$ and $t$ are different scalars), then $$\vec x_2 - \vec x_1 = (\vec x_0 + s\vec v) - (\vec x_0 + t\vec v) = (s-t)\vec v,$$ as desired.