Suppose $X\subset \Bbb P^n_{k}$ is a closed projective subscheme equidimensional of dimension $r$ with $k$ a field. Let $S(X)$ be the homogeneous coordinate ring of $X$, and $S(X)^{(d)}$ the $d^{th}$ graded piece. The Hilbert function $h_X$ is defined to send $d\mapsto \dim_{k} S(X)^{(d)}$ for nonnegative integers $d$. For $d$ large enough, $h_x$ is polynomial in $d$ of degree $r$ and this polynomial is called the Hilbert polynomial $p_X(d)$. The degree of $X$ is then defined to be $r!$ times the leading coefficient of the Hilbert polynomial.
One can prove that in the case of a finite set of distinct $k$-points, this gives exactly the same result as you are familiar with. But interesting things may happen with points over extension fields or thicker points.
Exercise: if a point is defined over an extension field $K\supset k$, then it's degree as a closed subscheme is exactly the degree of the field extension.
Exercise: if we define $tZ$, a $t$-thick point inside $\Bbb P^n_k$ supported at $Z$ to be of the form $I(Z)^t$ for $t>1$, then the degree of $tZ$ is $\binom{t+n-1}{n}$. (It's worth noting that there are fat/thick points not of this form, but this terminology is what's currently accepted in the field - people usually mean the $t^{th}$ infinitesimal neighborhood of the given point unless they say otherwise.)
Exercise: Degrees of finite sets of points add over disjoint unions. With this, you should have enough to come to the conclusions you're looking for.
You are correct that if $f : X \to Y$ is proper over some base $S$ then it has the property you described. If $Z \subset Y$ is a closed subscheme proper over $S$ then $f^{-1}(Z) \subset X$ is a closed subscheme proper over $S$.
Because $f^{-1}(Z)$ has $f^{-1}(Z)$ has a natural subscheme structure as $f^{-1}(Z) = Z \times_Y X$ which is not always reduced, it is probably better to require that $f^{-1}(Z)$ be proper with this standard subscheme structure.
However, it's also true if we give $f^{-1}(Z)$ the reduced subscheme structure because this is a closed subscheme of $f^{-1}(Z)$ with its standard structure and closed subschemes of proper schemes are proper.
Now let me prove the claim. The map $f^{-1}(Z) \to Z$ is proper because it is the base change of $f : X \to Y$. Therefore the composition $f^{-1}(Z) \to Z \to S$ is proper showing that $f^{-1}(Z)$ is proper over $S$.
However, I do not believe this property is enough to characterize proper morphisms even for varieties over an algebraically closed field. The problem is ``there aren't enough proper closed subvarieties''. For example, $\mathbb{A}^1 \setminus \{ 0 \} \to \mathbb{A}^1$ is not proper but the only proper closed subvarities of $\mathbb{A}^1$ are points whose preimages are also points.
In fact, we can make examples where $f : X \to Y$ is a closed surjective map of varieties over $\mathbb{C}$ with your property but not proper. For example, let $Y$ be the affine nodal curve $y^2 = x^2(x-1)$ and $X = \mathbb{A}^1 \setminus \{ i \}$. The normalization map $\nu : \tilde{Y} \to Y$ where $\tilde{Y} = \mathbb{A}^1$ sends $t \mapsto (t^2 + 1, t(t^2 + 1))$ so take $X \to \tilde{Y} \to Y$ i.e. we remove one of the two preimages of the node. Then $f$ is closed and surjective (even bijective!) pulling back proper closed subvarities (points) to proper closed subvarities. However, $f$ is not proper. To see this, consider $\tilde{f} : X \times \tilde{Y} \to Y \times \tilde{Y}$ and the closed set
$$\Delta = \{ (x,x) \mid x \in X \} \subset X \times \tilde{Y} $$
however $\tilde{f}(\Delta) = \{ (f(x), x) \mid x \in X \} \subset Y \times \tilde{Y}$ is the graph of $f$ minus one point and thus not closed.
However, for smooth varieties over $\mathbb{C}$, we can view an algebraic map $f : X \to Y$ as a map of complex manifolds $f^{\mathrm{an}} : X^{\mathrm{an}} \to Y^{\mathrm{an}}$ viewing $X, Y$ as complex manifolds with the analytic topology. Then it is a comforting fact that $f$ is proper iff $f^{\mathrm{an}}$ is proper in the usual topology sense.
Best Answer
(1) Let $ X $ be any scheme of finite type over a field $ k $. Let $ p \in X $ be a closed point. Choose an affine open $ U = \operatorname{Spec} A \subset X $ containing $ p $, then $ p $ is a closed point in $ U $ and hence corresponds to a maximal ideal $ \mathfrak{m} $ of $ A $. Since $ A $ is a finitely generated $ k $-algebra, so is the residue field $ k(\mathfrak{m}) = A/\mathfrak{m} $. By Zariski's lemma/Noether normalization, $ k(\mathfrak{m}) $ is a finite field extension of $ k $. Apply this to $ X = G = Gr(2,4, k) $ (I write the base field here because it's important) which is indeed of finite type - this is how you are defining the field of definition of a closed point. Indeed you should think of closed points as actual points, just that these points may not have 'coordinates' in $ k $.
(2) For any scheme $ X $ and a point $ p $ (not necessarily closed) and with residue field $ k(p) $, there is a canonical morphism $$ \operatorname{Spec} k(p) \rightarrow X $$ (A hint if you do not know this: check this when $ X $ is affine first and reduce the general case to the affine case by showing it is independent of the affine neighborhood you choose) Anyway, since your closed point has field of definition $ L $, there is a morphism $ \epsilon: \operatorname{Spec} L \rightarrow Gr(2,4, k) $. By the universal property of the Grassmanian, there is a universal rank $ 2 $ quotient $ \mathcal{G} $ on $ G $, $$ \mathcal{O}_G^4 \rightarrow \mathcal{G} \rightarrow 0 $$and pulling it back via $ \epsilon $ gives you a rank $ 2 $ quotient on $ \operatorname{Spec} L $,$$ \mathcal{O}_L^4 \rightarrow \mathcal{V} \rightarrow 0 $$ where $ \mathcal{V} $ now just corresponds to a rank $ 2 $ vector subspace of $ L^4 $. Projectivizing the last morphism shows $$ \mathbb{P} (\mathcal{V}) \hookrightarrow \mathbb{P} (\mathcal{O}_L^4) = \mathbb{P}^3_L $$ and this is the required line, viewed now as a closed subscheme of projective space over $ L $. This answers 3.
Bottom line: all of this is just a rigorous way of saying that your line may be living in a bigger field than $ k $. This is especially important in number theory. Hope it helps.