Let $A$ be a commutative ring, and $\operatorname{Spec} A$ be the spectrum of $A$ with its Zariski topology. A base for this topology is given by distinguished opens $U_f$. I am trying to verify that the presheaf on the base $U_f\mapsto A_f$ (the localization of $A$ by the multiplicative closure of $\{f\}$) satisfies the sheaf on a base axioms. I asked a question about sheaf axiom $1$ here, but I am now confused about to verify the second axiom.
My attempt is modeled off of Vakil's proof in the Geometry of schemes when $U=\operatorname{Spec} A$. Let $U_g$ be a distinguished open covered by (finitely many) $U_{f_i}$, and let $s_i\in A_{f_i}$ be sections which agree on overlaps. Now for some large enough $K$, we can regard $f_i^Ks_i$ as an element of $A$, $A_g$, or $A$, as if: $$s_i=\frac{a_i}{f_i^{n_i}}$$ then when $K\geq a_i$ we have that: $$f_i^Ks_i=\frac{f_i^{K-n_i}a_i}{1}$$ so the element $f_i^{K-n_i}$ in $A$, and the element $f_i^{K-n_i}/1$ in $A_g$ both map to $f_i^Ks_i$ in $A_f$. Now since $s_i$ and $s_j$ agree on overlaps, we have that $(f_if_j)^Ks_i$ and $(f_if_j)^Ks_j$ agree on overlaps, and taking $(f_if_j)^Ks_i$ and $(f_if_j)^Ks_j$ as elements of a ring, we have that for some larger $K$: $$(f_if_j)^Ks_j=(f_if_j)^Ks_i$$ We define $h_i\in A_g$ by taking $f_i^Ks_i$ as an element of $A_g$ and setting $h_i=f_i^Ks_i$. Then we have that: $$f_j^Kh_i=f_i^Kh_i$$ Now note that we for some $m$ we have that: $$g^m=\sum_{i}e_if_i^K$$ where $e_i\in A$. So we define the section of $U_g$, $s$ by: $$s=\sum_ie_ih_i$$ We would like this to agree on overlaps, but note that $s$ has the form $a/1$ for some $a\in A$, hence: $$s|_{U_j}= \frac{\sum_ie_ih_i}{1}$$ However, we have that as elements of $A$: $$f_j^Ks=\sum_i f_j^Ke_ih_i=\sum_ie_if_i^Kh_j=g^mh_j=g^mf_j^Ks_j$$
From here I am stuck, I have no idea what this $g^m$ factor is doing here or how to get it as some power of $f_j$ on both sides. In Vakil's proof, he gets $g^m=1$ because $A_g=A$, so it just works out. I suspect there is something I am misunderstanding about the proof, but my attempts to brute force it have left me in a barrage of indices that is borderline unreadable...any help would be greatly appreciated.
Ok I figured it out I think.
Let $U_g$ be a distinguished open and $U_{f_i}$ a finite open cover. Let $s_i\in A_{f_i}$ such that $$\begin{align*} \theta^i_{ij}(s_i)=\theta^j_{ij}(s_j) \end{align*}$$ Set: $$\begin{align*} s_i=\frac{a_i}{f_i^{l_i}} \end{align*}$$ and note that since $U_{f_if_j}\subset U_{f_i}, U_{f_j}$ we have that: $$\begin{align*} (f_if_j)^{k^i_{ij}}= b_{ij}^i f_i\qquad \text{and}\qquad (f_if_j)^{k^j_{ij}}= b_{ij}^j f_j \end{align*}$$ However, take $k^i_{ij}=k^j_{ij}=1$, then we have that $b^i_{ij}=f_j$, and $b^j_{ij}=f_i$. It follows that: $$\begin{align*} \theta^i_{ij}(s_i)=\frac{a_i\cdot f_j^{l_i}}{(f_if_j)^{l_i}}= \frac{a_j\cdot f_i^{l_j}}{(f_if_j)^{ l_j}}=\theta^j_{ij}(s_j) \end{align*}$$ so there exists some $K$ such that for all $i,j$: $$\begin{align*} (f_if_j)^K\cdot\left((f_if_j)^{ l_j}\cdot a_i\cdot f_j^{l_i} - (f_if_j)^{l_i}\cdot a_j\cdot f_i^{l_j}\right)=0 \end{align*}$$ multiply by $f_i^{l_i}f_j^{l_j}$ to obtain: $$\begin{align*} (f_if_j)^{K}\left( a_i\cdot f_j^{l_j} - a_j\cdot f_i^{l_i}\right)=0 \end{align*}$$ where $K=K+l_i+l_j$. Set $K$ to be large enough so that this expression holds for all $i$ and $j$, and define: $$\begin{align*} h_i=a_i f_i^{K} \end{align*}$$ Then note that for some $M$: $$\begin{align*} g^M=\sum_{i}e_i f_i^{K} \end{align*}$$ so set: $$ \begin{align*} s=\sum_i \frac{e_i h_i}{g^M} \end{align*}$$ and note that we have $f^{n_j}_j=bg$ for some $b$ and $n_j$. The restriction is then: $$\begin{align*} s|_{U_j}=\sum_i\frac{e_i \cdot h_i b^{M}}{f^{n_j M}} \end{align*}$$ We claim this equal to $s_j$; indeed we have that: $$\begin{align*} f^{K}_j\left(\sum_i{e_i \cdot h_i\cdot b}\cdot f^{l_i}_j-a_j\cdot f^{n_j M} \right) \end{align*}$$ Examine the first term: $$ \begin{align*} \sum_i{e_i \cdot h_i\cdot b^M}\cdot f^{K}_j\cdot f^{l_i}_i=& \sum_i{e_i \cdot f_i^{K}\cdot a_i \cdot b}\cdot f^{K}_j\cdot f^{l_i}_i \end{align*}$$ For each term we have that $f_i^Kf_j^Kf_i^{l_i}a_i=f_i^Kf_j^K f_j^{l_j}a_j$, hence: $$\begin{align*} \sum_i{e_i \cdot h_i\cdot b^M}\cdot f^{K}_j\cdot f^{l_i}_i=&(f_j^{l_j}f_j^Ka_jb^M)\sum_i f_i^Ke_i\\ =&(f_j^{l_j}f_j^Ka_jb^M)\cdot g^M\\ =&f_j^K\left(f_j^{l_j}a_j f^{n_jM}\right) \end{align*}$$ implying the claim.