Linear Equivalence of Divisors Implies Algebraic Equivalence The highlighted sections are taken  from Volume 1 of "Basic Algebraic Geometry" by Igor Shafarevich.

Definition:  Let $ X $ and $ T $ be two arbitrary irreducible varieties. For any point $ t \in T $ the map $ j_{t}: x \mapsto (x,t) $ defines an embedding $ X \hookrightarrow X \times T. $ Every divisor $ C $ on $ X \times T $ with $ \operatorname{Supp}C \not \supset X \times t $ defines a pullback divisor $ j_{t}^{*}(C) $ on $ X. $ In this case we say that $ j^{*}_{t}(C) $ is defined.

On page 188, the author shows that linear equivalence of divisors implies algebraic equivalence. There are a few details that I am having some difficulty with:

It is enough to prove this for equivalence to $ 0. $

I don't quite see why.
He proceeds with the proof: 

Suppose that $ D \in \operatorname{Div}X $ and $ D \sim 0, $ that is $ D = \operatorname{div}g $ with $ g \in k(X). $ Consider the variety $ T = \mathbb{A}^{2} \backslash (0,0), $ and write $ u,v $ for coordinates on $ \mathbb{A}^{2}. $ We can view $ g,u,v $ as functions on $ X \times T, $ meaning the pullbacks $ p^{*}(g),q^{*}(u) $ and $ q^{*}(v), $ where as usual $ p: X \times T \rightarrow X $ and $ q: X \times T \rightarrow T $ are the projections. 

In what sense does invoking the pullbacks equate to viewing $ g,u $ and $ v $ as functions on $ X \times T$?
And finally:

Set $ C = \operatorname{div}(u+vg) $ and consider the algebraic family of divisors $ f $ defined by the divisor $ C $ on $ X \times T. $ One checks that $ f(1,0) = 0$(the zero divisor) and $ f(0,1) = D, $ and hence $ D \equiv 0. $

I understand the reason why we are checking that $ f(0,1) = D, $ and $ f(0,1) = 0. $ By definition of algebraic equivalence, we only need to check that there are such points in $ T = \mathbb{A}^{2}\backslash (0,0) $ whose images are the divisors in question. My problem is that I don't know how to actually check this. I know that $ f(t) = j^{*}_{t})(C) $ by definition of a family of divisors, but I don't know how to use this.
 A: 
It is enough to prove this for equivalence to 0.

In the paragraph immediately before this, Shafarevich notes that algebraic equivalence is compatible with addition. So $D_1 \equiv D_2$ if and only if $D_1 - D_2 \equiv 0$.

In what sense does invoking the pullbacks equate to viewing $ g,u $ and $ v $ as functions on $ X \times T$?

An element $g \in k(X)$ of the function field is the same as a rational map $g: X \dashrightarrow \mathbb{P}^1$. The pullback map $p^*: k(X) \to k(X \times T)$ is just precomposition, so $p^*(g) = g \circ p: X \times T \to X \dashrightarrow \mathbb{P^1}$.

Set $ C = \operatorname{div}(u+vg) $ and consider the algebraic family of divisors $ f $ defined by the divisor $ C $ on $ X \times T. $ One checks that $ f(1,0) = 0$(the zero divisor) and $ f(0,1) = D, $ and hence $ D \equiv 0. $

Since $u, v$ are the coordinates on $T = \mathbb{A}^2 \setminus \{(0,0)\}$, he's just saying set $(u,v) = (1,0)$ and $(u,v) = (0,1)$. When $(u,v) = (1,0)$ we have $C|_{(u,v) = (1,0)} = \DeclareMathOperator{\div}{div} \div(1) = 0$ since the constant function $1$ has no zeroes or poles, and $C|_{(u,v) = (0,1)} = \div(g) = D$.
