Adjunction formula (Griffiths & Harris proof)

The normal bundle
Choose on each $U_i$ a coordinate system $(z_1^{i},...,z_n^{i})$ such that $U_i\cap V$ is given by the equation $z_n^{i}=0$, so that $f_i=z_n^{i}$ .
The normal bundle $N_V$ on $V \;$ is then given by the cocycle

$$n_{ij}=\frac {\partial z_n^{i}}{\partial z_n^{j}} \in \mathcal O^*(V \cap U_{ij})$$ Note carefully that this bundle is defined only on $V$ , and not on $M$.

The bundle associated to $V$
In the same notation the bundle $\mathcal O(V)=[V]$ is defined by a cocycle $g_{ij}\in \mathcal O^*(U_{ij})$ satisfying
$z^i_n=z^j_n.g_{ij}$ on $U_{ij}$.
Taking partial derivatives with respect to $z^j_n$ yields $\frac {\partial z_n^{i}}{\partial z_n^{j}} = g_{ij}+z^j_n.\frac {\partial g_{ij}}{\partial z^j_n} $.
Restricting this to $V \cap U_{ij}$, we get a cocycle for $[V]|V$ ( remember that $z^j_n=0$ on $V\cap U_j$) $$\frac {\partial z_n^{i}}{\partial z_n^{j}} = g_{ij} \in \mathcal O^*(V \cap U_{ij})$$

The two displayed equations prove that $n_{ij}=g_{ij}$ and thus that $$N_V \simeq[V]|V$$

(As you see, I prefer to avoid differential forms and dualization : no $N_V^*$, no $[-V]$. )