Question in Proposition $1.40$ of Algebraic Geometry $1$ by Gortz and Wedhorn

I am unable to understand the proof of proposition $1.40$ given on Page $21$ of the textbook by Gortz and Wedhorn.

Definition $1.30$ Let $X\subset \mathbb{A}^n{k}$ be the affine algebraic set The $k-$algebra $\Gamma(X)= k[T_1,...,T_n]\cong Hom (X, \mathbb{A}^1(k))$ is called the affine coordinate ring of $X$.

Definition $1.39$ Let $X$ be an irreducible affine algebraic set and $\phi \neq U \subseteq X$ be open . Denote by $m_x$ the maximal ideal of $\Gamma(X)$ corresponding to $x\in X$ and by $\Gamma(X)_{m_x}$ the localization of affine coordinate ring wrt to $m_x$. We define $O_X(U) =\cap_{x\in U} \Gamma(X)_{m_x}$. We call $(X,O_X)$ the space with functions associated with $X$.

Proposition $1.40$ Let $(X,O_X)$ be the space with functions associated to the irreducible affine algebraic set $X$ and let $f\in \Gamma(X)$. Then there exists an equality $O_X(D(f))= \Gamma(X)_f$.

Proof: Clearly we have $\Gamma(X)_f\subset O_X(D(f))$ ( I am not sure why it should hold , can you please explain?). Let $g \in O_X((D(f))$ and set $a=${$h\in \Gamma(X); hg \in \Gamma(X)$}. $a$ is an ideal of $\Gamma(X)$ and we have to show that $f\in rad(a)$(Why we need to show this?) . By Nullstellansatz , we have $rad{a}=I(V(a))$ so it's sufficient to show that $f(x)=0$ for all $x\in V(a)$. Let $x\in X$ be a point with $f(x)\neq0$ ie $x\in D(f)$. As $g\in O_X(D(f))$, we have $g_1 , g_2 \in \Gamma(X)$ , $g_2 \notin m_x$ with $g=\frac{g_1}{g_2}$. Thus $g_2 \in a$ (How?)and as $g_2(x)\neq 0$ we have $x\notin V(a)$.

Please help me with questions in this proof. I have no one in real life to ask these as I am from a poor nation.