Suppose that your eye is at the origin, and the "canvas" on which you draw is in front of your eye, and is the $z = 1$ plane. Then the point $(2, 3, 5)$ in space will project, in your "drawing" or "seeing" of the world, to the point $(2/5, 3/5, 1)$ in the $z = 1$ plane. In general, any point $(x, y, z)$ will project to $(x/z, y/z, 1)$.
Now consider two parallel lines; the first one consisting of all points of the form $(1, t, 4)$ and the second consisting of points $(3, t, 2)$. The projections of these to the drawing plane will consist of points of the form $(1/4, t/4, 1)$ and $(3/2, t/2, 1)$, respectively, i.e., they'll still be parallel.
Now look at the lines $(-1, 0, t)$ and $(1, 0, t)$. These project to
$(-1/t, 0, 1)$ and $(1/t, 0, 1)$, which "meet" at the point $(0, 0, 1)$ when $t$ goes to $\infty$.
Does that help at all?
Really reasonable questions.
The flat limit is by definition the fiber of the closure of $\Psi$ (note that this is true scheme theoretically not just set theoretically).
So to show that the flat limit is contained in something, one way is to exhibit an appropriate closed set containing $\Psi$, more precisely to exhibit the fiber of such a set. In this case take the set
$$\Psi’ = \{(t,L) \in \mathbb{A}^1 \times \mathbb{G}(1,3) : L \cap C_t \ne \emptyset \}$$
where $C_t$ is $A_t(C)$ for $t\ne0$ and is the flat limit of that curve at $t=0$. Note that this definition also dropped the $t\ne 0$ requirement, so it’s almost like a “guess” as to what the closure of $\Psi$ is.
This gives a closed set (since the total space of the flat family $C_t$ is closed in $\mathbb{A}^1 \times \mathbb{P}^3$), so it contains the closure of $\Psi$. And the fiber of it is set-theoretically the set of lines meeting $C_0$.
Next: is it true that a flat limit of divisors is a divisor?
Yes. First of all fiber dimension is constant in flat families. But more concretely, the closure of $\Psi$ is a divisor on $\mathbb{A}^1 \times \mathbb{G}(1,3)$, so restricting it to $t=0$ gives a divisor on $\mathbb{G}(1,3)$. (In general the cycle class is also preserved under flat limits over $\mathbb{P^1}$.)
As for multiplicity. One nice thing here is that divisors on smooth varieties are Cohen-Macaulay, so they are reduced iff generically reduced. So, no embedded points. In this case the key fact is that $C_0$ is reduced away from $q$. This is actually a consequence of the fact that $C$ is transverse to $H$ at each of the $p_i$’s. As $t\to 0$, the curve “stretches” away from $H$ and flattens into a line through $p_i$. (If $C$ were tangent to $H$ at $p_i$ it would become a double line, etc.)
The local model of this would be to take $C$ itself to be a line passing through $H$ at just one point, in which case the calculation is (I think) easy to do directly.
Best Answer
a) Indeed there is a very classical duality between points and lines in projective space $\mathbb P^2$.
Points $P=(x:y:z)\in \mathbb P^2$ live in one projective plane and lines $l\subset \mathbb P^2$, which are subsets of that plane, can also been seen as points in a new projective space $ { \mathbb P^2}^*$.
To the line $l\subset \mathbb P^2$ with equation $ax+by+cz=0$ one just associates the point $ [l]=(a:b:c)\in { \mathbb P^2}^*$ .
b) To get back to your question: if you fix a point $P_0=(x_0:y_0:z_0)\in \mathbb P^2$, a line $l$ given by $ax+by+cz=0 $ will go through $P_0$ if and only if $ax_0+by_0+cz_0=0$.
In the dual perspective the set of all the corresponding points $[l]\in { \mathbb P^2}^*$ will describe a line in the new plane: $L_{P_0}\subset { \mathbb P^2}^*$ called, as you correctly stated, a pencil of lines.
The equation of $L_{P_0}$ is $x_0a+y_0b+z_0c=0$, where now $a, b $ and $c$ are seen as variables!
Concretely: if $P_0=(1:2:-1)$, the pencil $L_{P_0}$ consists of the lines $[l]\in { \mathbb P^2}^*$whose equation $ax+by+cz=0$ satisfies $a+2b-c=0$.
So the line $3x+4y+11z=0$ belongs to the pencil but the line $x-2y+4z=0$ does not.
c) This is the beginning of the theory of duality of projective spaces, an intensively studied subject in algebraic geometry, created at the beginning of the 19th century by two officers in the French army, Gergonne and Poncelet, who quarreled bitterly over the priority of this discovery...