Let $X$ be a complex manifold , $L$ a holomorphic line bundle over $X$ whose transition denoted by {$U_i,g_{i,j}$} , and $s$ be a nontrivial holomorphic section of $L$ which is a collection of holomorphic functions {$s_i$}. Assume $S=s^{-1}(0)$ is a smooth hypersurface. By some simple calculations, we deduce $$ds_i=g_{i,j}ds_j$$ on $U_i\cap U_j\cap S$. Then someone claim that $ds$ is a $L$-valued $(1,0)$-form on $S$.
However, the operator $d$ belongs to $X$. More precisely, let {$z_1,...,z_n$} be a local coordinate of $X$ around $S$, then $$ds_i=\Sigma_k\frac{\partial s_i}{\partial z_k}dz_k .$$ If locally $s_i=z_n$, then $ds_i=dz_n$. However, $dz_n$ is not a form on the manifold $S$ whose holomorphic cotangent bundle is spanned by{ $dz_1,...dz_{n-1}$}.
My question: What is $ds$ on earth? It is a bundle-valued form on $S$ or a bundle-valued form on $X$ restricted to $S$ or something else?
It is indeed a restriction to $S$ of a form on $X$ with values on $L$ i.e. $ds$ defines a section of $(\Omega_X^1\otimes L)|_S$.
Let $\iota \colon S \rightarrow X$ be the inclusion. Note that $\iota^\ast ds_j =0$ since the pullback commutes with $d$ and $S\cap U_j=\{s_j=0\}$ so it is zero as a form on $S$. However we can use $ds$ to show that $L|_S$ is the normal bundle of $S$.
On the one hand the pullback of one-forms leads to an exact sequence $$ 0 \rightarrow N_S^\ast \rightarrow \Omega_X^1|_S \xrightarrow{\iota^\ast} \Omega^1_S \rightarrow 0 $$ $N_S$ being the normal bundle of $S$ on $X$. On the other hand $ds$ defines a map $L^\ast|_S \hookrightarrow \Omega_X^1|_S$ that falls into the kernel of $\iota^\ast$. Hence $ds$ induces a map of line bundles $L^\ast|_S \rightarrow N_S^\ast$ which is an isomorphism. Indeed this map is locally given by multiplication with $ds_j$ that does not vanish on $S$.