[proofplan] We prove the adjunction isomorphism by writing both line bundles in local coordinates adapted to the smooth divisor. On each coordinate neighbourhood where $D$ is cut out by one holomorphic coordinate, we build a local frame of $\left(K_X \otimes \mathcal{O}_X(D)\right)|_D$ and compare its transition functions with those of $K_D$. The determinant calculation shows that the transition functions agree, so the local identifications glue to a natural global isomorphism. Finally, when $D$ is the zero divisor of a section of $L$, the section identifies $L$ with $\mathcal{O}_X(D)$. [/proofplan]

[step:Choose local coordinates adapted to the smooth divisor]Let $\{U_a\}_{a \in A}$ be an open cover of $X$ such that, for every $a \in A$, there is a holomorphic coordinate chart \begin{align*} z_a: U_a &\to z_a(U_a) \subset \mathbb{C}^n, \\ p &\mapsto \left(z_{a,1}(p),\dots,z_{a,n}(p)\right), \end{align*} with \begin{align*} D \cap U_a = \{p \in U_a : z_{a,n}(p)=0\}. \end{align*} Write \begin{align*} V_a := D \cap U_a. \end{align*} Then \begin{align*} x_a: V_a &\to \mathbb{C}^{n-1}, \\ p &\mapsto \left(z_{a,1}(p),\dots,z_{a,n-1}(p)\right) \end{align*} is a holomorphic coordinate chart on $D$. For each $a \in A$, define the local holomorphic $n$-form \begin{align*} \Omega_a := dz_{a,1} \wedge \cdots \wedge dz_{a,n} \end{align*} on $U_a$, and define the local holomorphic $(n-1)$-form \begin{align*} \omega_a := dx_{a,1} \wedge \cdots \wedge dx_{a,n-1} \end{align*} on $V_a$.[/step]

[guided]Because $D$ is a smooth divisor, near every point of $D$ it is cut out by one holomorphic coordinate. Thus we may cover $X$ by coordinate neighbourhoods $U_a$ on which the divisor is locally the hypersurface $z_{a,n}=0$. The remaining coordinates restrict to coordinates on $D$: \begin{align*} x_a(p) = \left(z_{a,1}(p),\dots,z_{a,n-1}(p)\right). \end{align*} The canonical bundle $K_X$ is locally generated by the top holomorphic form \begin{align*} \Omega_a = dz_{a,1} \wedge \cdots \wedge dz_{a,n}, \end{align*} while the canonical bundle $K_D$ is locally generated by \begin{align*} \omega_a = dx_{a,1} \wedge \cdots \wedge dx_{a,n-1}. \end{align*} The proof will compare the transition functions for these two kinds of frames after tensoring $K_X$ with the divisor line bundle $\mathcal{O}_X(D)$.[/guided]

[step:Compute the transition functions for $K_D$]On an overlap $V_a \cap V_b$, define the [holomorphic function](/page/Holomorphic%20Function) \begin{align*} A_{ab}: V_a \cap V_b &\to \operatorname{Mat}_{(n-1)\times(n-1)}(\mathbb{C}), \\ p &\mapsto \left(\frac{\partial z_{a,i}}{\partial z_{b,j}}(p)\right)_{1 \le i,j \le n-1}. \end{align*} Since $x_a$ and $x_b$ are coordinate systems on $D$, the change of coordinates on $D$ gives \begin{align*} \omega_a = \det(A_{ab})\, \omega_b \end{align*} on $V_a \cap V_b$.[/step]

[guided]On $V_a \cap V_b$, both $x_a$ and $x_b$ are holomorphic coordinate systems for the same complex manifold $D$. The Jacobian matrix of this coordinate change is \begin{align*} A_{ab}(p)=\left(\frac{\partial z_{a,i}}{\partial z_{b,j}}(p)\right)_{1 \le i,j \le n-1}. \end{align*} This matrix records only the tangential coordinates, because on $D$ the coordinate $z_{a,n}$ is normal to the divisor. The transformation law for a wedge product of differentials gives \begin{align*} dx_{a,1} \wedge \cdots \wedge dx_{a,n-1} = \det(A_{ab})\, dx_{b,1} \wedge \cdots \wedge dx_{b,n-1}. \end{align*} Therefore \begin{align*} \omega_a = \det(A_{ab})\,\omega_b. \end{align*} So $\det(A_{ab})$ is the transition function of $K_D$ from the frame $\omega_b$ to the frame $\omega_a$.[/guided]

[step:Compute the transition functions for $\left(K_X \otimes \mathcal{O}_X(D)\right)|_D$]For each $a \in A$, let \begin{align*} f_a: U_a &\to \mathbb{C}, \\ p &\mapsto z_{a,n}(p) \end{align*} be the local defining function of $D$ on $U_a$. Let $e_a$ denote the local frame of $\mathcal{O}_X(D)$ on $U_a$ corresponding to the meromorphic section $1/f_a$. On $U_a \cap U_b$, there is a nowhere-vanishing holomorphic function \begin{align*} u_{ab}: U_a \cap U_b &\to \mathbb{C}^\times \end{align*} such that \begin{align*} f_a = u_{ab} f_b. \end{align*} Hence \begin{align*} e_a = u_{ab}^{-1} e_b. \end{align*} Let \begin{align*} J_{ab}: U_a \cap U_b &\to \mathbb{C}, \\ p &\mapsto \det\left(\frac{\partial z_{a,i}}{\partial z_{b,j}}(p)\right)_{1 \le i,j \le n}. \end{align*} Then \begin{align*} \Omega_a = J_{ab}\,\Omega_b. \end{align*} Since $f_a=u_{ab}f_b$, differentiating in the normal coordinate direction and restricting to $D$ gives \begin{align*} \frac{\partial z_{a,n}}{\partial z_{b,n}}\bigg|_D = u_{ab}\big|_D, \end{align*} while differentiating along tangent directions gives \begin{align*} \frac{\partial z_{a,n}}{\partial z_{b,j}}\bigg|_D = 0 \end{align*} for $1 \le j \le n-1$. Therefore, along $D$, \begin{align*} J_{ab}\big|_D = \det(A_{ab})\, u_{ab}\big|_D. \end{align*} It follows that \begin{align*} (\Omega_a \otimes e_a)\big|_D = \det(A_{ab})\,(\Omega_b \otimes e_b)\big|_D. \end{align*}[/step]

[guided]The divisor line bundle $\mathcal{O}_X(D)$ is locally generated by the meromorphic section with one simple pole along $D$. In the chart $U_a$, the divisor is defined by \begin{align*} f_a(p)=z_{a,n}(p), \end{align*} so a local frame of $\mathcal{O}_X(D)$ is denoted by $e_a=1/f_a$. On an overlap $U_a \cap U_b$, the two defining functions define the same smooth divisor, so they differ by a nowhere-vanishing holomorphic unit: \begin{align*} f_a = u_{ab} f_b, \end{align*} where \begin{align*} u_{ab}: U_a \cap U_b \to \mathbb{C}^\times \end{align*} is holomorphic. Therefore \begin{align*} e_a = \frac{1}{f_a} = \frac{1}{u_{ab}f_b} = u_{ab}^{-1}e_b. \end{align*} Now compare the canonical bundle transition functions. Define \begin{align*} J_{ab}(p)=\det\left(\frac{\partial z_{a,i}}{\partial z_{b,j}}(p)\right)_{1 \le i,j \le n}. \end{align*} Then \begin{align*} \Omega_a = J_{ab}\Omega_b. \end{align*} The key point is that $J_{ab}$ splits into a tangential determinant and a normal factor along $D$. Since \begin{align*} z_{a,n}=f_a=u_{ab}f_b=u_{ab}z_{b,n}, \end{align*} we differentiate with respect to $z_{b,j}$. For $1 \le j \le n-1$, \begin{align*} \frac{\partial z_{a,n}}{\partial z_{b,j}} = \frac{\partial u_{ab}}{\partial z_{b,j}}z_{b,n} + u_{ab}\frac{\partial z_{b,n}}{\partial z_{b,j}} = \frac{\partial u_{ab}}{\partial z_{b,j}}z_{b,n}, \end{align*} and this restricts to $0$ on $D$ because $z_{b,n}=0$ there. For the normal direction, \begin{align*} \frac{\partial z_{a,n}}{\partial z_{b,n}} = \frac{\partial u_{ab}}{\partial z_{b,n}}z_{b,n} + u_{ab}, \end{align*} so restriction to $D$ gives \begin{align*} \frac{\partial z_{a,n}}{\partial z_{b,n}}\bigg|_D = u_{ab}\big|_D. \end{align*} Thus the full Jacobian matrix is block upper triangular along $D$, with tangential block $A_{ab}$ and normal entry $u_{ab}|_D$. Hence \begin{align*} J_{ab}\big|_D = \det(A_{ab})\,u_{ab}\big|_D. \end{align*} Combining this with $e_a=u_{ab}^{-1}e_b$, we obtain \begin{align*} (\Omega_a\otimes e_a)\big|_D = (J_{ab}\Omega_b \otimes u_{ab}^{-1}e_b)\big|_D = \det(A_{ab})(\Omega_b\otimes e_b)\big|_D. \end{align*} This is exactly the same transition function as the one computed for $K_D$.[/guided]

[step:Glue the local identifications into the adjunction isomorphism] For every $a \in A$, define a holomorphic line bundle map over $V_a$ \begin{align*} \Phi_a: K_D|_{V_a} &\to \left(K_X \otimes \mathcal{O}_X(D)\right)|_{V_a}, \\ \omega_a &\mapsto (\Omega_a \otimes e_a)|_{V_a}. \end{align*} Both $\omega_a$ and $(\Omega_a \otimes e_a)|_{V_a}$ are local frames, so $\Phi_a$ is an isomorphism on $V_a$. On $V_a \cap V_b$, the previous two steps show that \begin{align*} \omega_a = \det(A_{ab})\omega_b \end{align*} and \begin{align*} (\Omega_a \otimes e_a)|_D = \det(A_{ab})(\Omega_b \otimes e_b)|_D. \end{align*} Therefore $\Phi_a=\Phi_b$ on $V_a \cap V_b$. The maps $\Phi_a$ glue to a global holomorphic line bundle isomorphism \begin{align*} \Phi: K_D &\to \left(K_X \otimes \mathcal{O}_X(D)\right)|_D. \end{align*} This proves \begin{align*} K_D \cong \left(K_X \otimes \mathcal{O}_X(D)\right)|_D. \end{align*} [/step]

[step:Identify $\mathcal{O}_X(D)$ with $L$ when $D$ is the zero divisor of a section] Assume now that $L \to X$ is a holomorphic line bundle and that $s \in H^0(X,L)$ has zero divisor $D$. Choose, on each $U_a$, a holomorphic frame \begin{align*} \ell_a: U_a &\to L|_{U_a} \end{align*} such that \begin{align*} s = f_a \ell_a. \end{align*} Define a local holomorphic bundle map \begin{align*} \Psi_a: \mathcal{O}_X(D)|_{U_a} &\to L|_{U_a}, \\ e_a &\mapsto \ell_a. \end{align*} On $U_a \cap U_b$, the equality $f_a=u_{ab}f_b$ and the identity $s=f_a\ell_a=f_b\ell_b$ imply \begin{align*} \ell_a = u_{ab}^{-1}\ell_b. \end{align*} Since $e_a=u_{ab}^{-1}e_b$, the maps $\Psi_a$ agree on overlaps. Hence they glue to a holomorphic line bundle isomorphism \begin{align*} \Psi: \mathcal{O}_X(D) &\to L. \end{align*} Restricting to $D$ and tensoring with $K_X|_D$ gives \begin{align*} \left(K_X \otimes \mathcal{O}_X(D)\right)|_D \cong \left(K_X \otimes L\right)|_D. \end{align*} Combining this with the adjunction isomorphism already proved yields \begin{align*} K_D \cong \left(K_X \otimes L\right)|_D. \end{align*} [/step]