I am reading Griffiths and Harris book Principles of Algebraic geometry.
Let $L\rightarrow M$ be a holomorphic line bundle, with trivializations $\varphi_\alpha:\pi^{-1}(U_\alpha)\rightarrow U_\alpha\times \mathbb{C}$ over an open cover $\{U_\alpha\}$ of $M$ and transition functions $\{g_{\alpha\beta}\}$ relative to $\{\varphi_\alpha\}$.
Given this set up, It says the following:
Any section of $L$ over $U\subseteq M$ is given exactly by a collection if functions $s_\alpha\in\mathcal{O}(U\cap U_\alpha)$ satisfying $s_\alpha=g_{\alpha\beta}s_\beta$ in $U\cap U_\alpha\cap U_\beta$.
Just for simplicity, I would like to consider particular case :
Any section of $L$ over $M$ is given exactly by a collection of functions $s_\alpha\in\mathcal{O}(U_\alpha)$ satisfying $s_\alpha=g_{\alpha\beta}s_\beta$ in $U_\alpha\cap U_\beta$.
Given $\varphi_\alpha:\pi^{-1}(U_\alpha)\rightarrow U_\alpha\times \mathbb{C}$ we have a section $\sigma_\alpha:U_\alpha\rightarrow L$ given by $\sigma_\alpha(u)=\varphi_\alpha^{-1}(u,1)$. One would hope to glue these sections $\sigma_\alpha:U_\alpha\rightarrow L$ to get a section $\sigma:M\rightarrow L$. But the problem is there is no reason why would $\sigma_\alpha(x)=\sigma_\beta(x)$ for $x\in U_\alpha\cap U_\beta$. So, we should look for another collection of sections on $U_\alpha$ that we can glue.
We have $s_\alpha:U_\alpha\rightarrow \mathbb{C}$. This would give another section $s_\alpha\sigma_\alpha:U_\alpha\rightarrow L$ with $u\mapsto s_\alpha(u)\sigma_\alpha(u)$.
Now, the condition $g_{\alpha\beta}s_\beta=s_\alpha$ gives some hope to glue these sections $s_\alpha\sigma_\alpha$. Only thing we need to prove is that $s_\alpha\sigma_\alpha=s_\beta\sigma_\beta$ on $U_\alpha\cap U_\beta$ i.e., $s_\alpha(u)\sigma_\alpha(u)=s_\beta(u)\sigma_\beta(u)$ for each $u\in U_\alpha\cap U_\beta$.
https://mathoverflow.net/questions/297389/connection-on-a-principal-bundle-and-transition-functions-as-in-hitchins-notes/297409?noredirect=1#comment740855_297409 says that $\sigma_\beta=g_{\alpha\beta}\sigma_\alpha$.
We have $s_\alpha=g_{\alpha\beta}s_\beta$, which imply $\sigma_\alpha s_\alpha=\sigma_\alpha g_{\alpha\beta}s_\beta$ which imply that $\sigma_\alpha s_\alpha=\sigma_\beta s_\beta$ as $\sigma_\alpha g_{\alpha\beta}=\sigma_\beta$. Thus, we can glue sections $s_\alpha\sigma_\alpha$ to get a section $\sigma:M\rightarrow L$ such that $\sigma|_{U_\alpha}=s_\alpha\sigma_\alpha$.
Given a section $s:M\rightarrow L$ I want to construct $s_\alpha:U_\alpha\rightarrow \mathbb{C}$ such that $s_\alpha=g_{\alpha\beta}s_\beta$.
We already have sections $\sigma_\alpha:U_\alpha\rightarrow L$. Let $x\in U_\alpha$ then, we have $s(x)\in \pi^{-1}(x)$ and $\sigma_\alpha(x)\in \pi^{-1}(x)$. As $\pi^{-1}(x)$ is $1$ dimensional, there exists $a\in \mathbb{C}$ such that $s(x)=a\sigma_\alpha(x)$ and this $a$ is what I plan to declare as $s_\alpha(x)$. This gives $s_\alpha:U_\alpha\rightarrow \mathbb{C}$. It remains to see why would $s_\alpha=g_{\alpha\beta}s_\beta$ on $U_\alpha\cap U_\beta$.
For $x\in U_\alpha\cap U_\beta$, I have $s(x)=s_\alpha(x)\sigma_\alpha(x)$ and $s(x)=s_\beta(x)\sigma_\beta(x)$. This is same as saying $s_\alpha(x)\sigma_\alpha(x)=s_\beta(x)\sigma_\beta(x)$. As $\sigma_\beta=\sigma_\alpha g_{\alpha\beta}$ we have $s_\alpha(x)\sigma_\alpha(x)=s_\beta(x)\sigma_\beta(x) =s_\beta(x)\sigma_\alpha(x)g_{\alpha\beta}(x)$ i.e., we have $$s_\alpha(x)\sigma_\alpha(x)=s_\beta(x)g_{\alpha\beta}(x)\sigma_\alpha(x).$$ As $\sigma_\alpha(x)\neq 0\in \pi^{-1}(x)$ this would mean $s_\alpha(x)=s_\beta(x)g_{\alpha\beta}(x)$ i.e., $s_\alpha=s_\beta g_{\alpha\beta}$.
As $\psi_\alpha^{-1}(x,0)=0$ and $\psi_\alpha^{-1}$ is bijection, we can not have $\sigma_\alpha(x)=\psi^{-1}(x,0)$ equal to $0$.
Can some one please confirm if this is correct or are there any gaps.