Recall that for a line bundle $L$, we can take local trivialization $\phi:\pi^{-1}(U_\alpha)\to U_\alpha\times\mathbb C$. Take local frame (one section, since line bundle is of rank 1), say $s_\alpha=\phi^{-1}(\pi(\cdot),1)$, then any section of $L$ can be represented by $s|_{U_\alpha}=f_\alpha s_\alpha$, for some complex function $f_\alpha$. The transition functions are defined on $U_{\alpha\beta}:=U_\alpha\cap U_\beta$, as
$$
s_\alpha=g_{\alpha\beta}s_\beta.
$$
Thus, for the coordinates of $s$:
$$
f_\beta=g_{\alpha\beta}f_\alpha.
$$
Locally, a connection of $L$ can be writen as:
$$
\nabla=d+A_\alpha,
$$
where $A_\alpha$ are complex valued 1-form over $U_\alpha$.
In fact,
$$
[\nabla s]|_{U_\alpha}=\nabla(f_\alpha s_\alpha)=df_\alpha s_\alpha+f_\alpha\nabla s_\alpha,
$$
thus, if we define $\nabla s_\alpha:=A_\alpha s_\alpha$, then we get
$$
[\nabla s]|_{U_\alpha}=(df_\alpha+f_\alpha A_\alpha) s_\alpha,
$$
which is the precise mean when we write $\nabla=d+A_\alpha$.
We need to know the transition relation of $A_\alpha$. This can be compute as follows:
$$
A_\alpha g_{\alpha\beta}s_\beta=A_\alpha s_\alpha=\nabla s_\alpha=\nabla(g_{\alpha\beta} s_\beta)=dg_{\alpha\beta} s_\beta+g_{\alpha\beta}A_\beta s_\beta.
$$
Thus,
$$
A_\alpha g_{\alpha\beta}=dg_{\alpha\beta}+g_{\alpha\beta}A_\beta\implies
A_\beta=g_{\alpha\beta}^{-1}A_\alpha g_{\alpha\beta}-g_{\alpha\beta}^{-1}dg_{\alpha\beta}=A_\alpha-g_{\alpha\beta}^{-1}dg_{\alpha\beta}.
$$
To move on, I will use a equivalent definition of curvature: $F_\nabla=D^2$, where $D$ is the exterior differential. In particular, by the definition of $D$, we know that(locally)
\begin{align*}
D^2s&=D(\nabla s)=D[(d f_\alpha+f_\alpha A_\alpha)s_\alpha]\\
&=d(df_\alpha+A_\alpha f_\alpha)s_\alpha+(-1)^1(df_\alpha+A_\alpha f_\alpha)\wedge\nabla s_\alpha\\
&=[dA_\alpha f_\alpha-A_\alpha\wedge df_\alpha-(df_\alpha+A_\alpha f_\alpha)\wedge A_\alpha]s_\alpha\\
&=[dA_\alpha f_\alpha-A_\alpha\wedge df_\alpha+A_\alpha\wedge(df_\alpha+A_\alpha f_\alpha)]s_\alpha\\
&=[dA_\alpha+A_\alpha\wedge A_\alpha]f_\alpha s_\alpha\\
&=(dA_\alpha+A_\alpha\wedge A_\alpha)s.
\end{align*}
Therefore,
$$
F_\nabla|_{U_\alpha}=dA_\alpha+A_\alpha\wedge A_\alpha.
$$
Note that $A_\alpha$ is just a complex valued 1-form (in general it is a matrix valued 1-form), we have$A_\alpha\wedge A_\alpha=0$ and
$$
F_\nabla|_{U_\alpha}=dA_\alpha.
$$
Now, we are ready to show, $F_\nabla$ is a global defined 2-form. In fact,
\begin{align*}
F_\nabla|_{U_\beta}&=dA_\beta=d(A_\alpha-g_{\alpha\beta}^{-1}d g_{\alpha\beta})\\
&=dA_\alpha-dg_{\alpha\beta}^{-1}\wedge dg_{\alpha\beta}\\
&=dA_\alpha+g_{\alpha\beta}^{-1}dg_{\alpha\beta}g_{\alpha\beta}^{-1}\wedge dg_{\alpha\beta}\\
&=dA_\alpha+g_{\alpha\beta}^{-1}dg_{\alpha\beta}\wedge g_{\alpha\beta}^{-1}dg_{\alpha\beta}\\
&=dA_\alpha\\
&=F_\nabla|_{U_\alpha}.
\end{align*}
This answers your first question.
For the second, it is standard in Riemaniann geometry and I believe your can find in text book.
I would like to remark that, the above computation is trivial but maybe you need to do it once again by yourself.
You can use the fact that the pullback connection preserves parallel transport, in the sense that the pullback of a parallel transport is the same as the parallel transport of the pullback. You can then use the definition of curvature as the composition of parallel transports, which yields the same result. You are missing some pullbacks in your formula, however.
Here is an explicit calculation, where I'm using $\varphi$ instead of f
$
F^{\varphi^*\nabla}(X,Y)\varphi^*s=
\varphi^*\nabla_X \varphi^*\nabla_Y \varphi^*s - \varphi^*\nabla_Y \varphi^*\nabla_X \varphi^*s -\varphi^*\nabla_{[X,Y]}\varphi^*s= \\
\varphi^*\nabla_X \varphi^*(\nabla_{d \varphi (Y)}s)-\varphi^*\nabla_Y \varphi^*(\nabla_{d \varphi (X)}s)-\varphi^*(\nabla_{d \varphi([X,Y])}s)=\\
\varphi^*\Big(\nabla_{d\varphi(X)}\nabla_{d\varphi(Y)}s-\nabla_{d\varphi(Y)}\nabla_{d\varphi(X)}s-\nabla_{[d \varphi(X),d\varphi(Y)]}s\Big)= \varphi^*(F^\nabla(d \varphi(X),d
\varphi(Y))s)
$
And the last step uses the naturality of the Lie bracket.
Best Answer
Yes, if you use coordinate vector fields, then the Lie brackets all vanish and the formula simplifies to $$F(\partial_i, \partial_j)s = \nabla_{\partial_i}\nabla_{\partial_j}s - \nabla_{\partial_j}\nabla_{\partial_i}s.$$