[guided]The goal is to prove that every vector in the fibre $M|_x$ is the value at $x$ of some global holomorphic section of $M$. Let $M_{x,\mathrm{st}}$ denote the stalk at $x$ of the sheaf of holomorphic sections of $M$. The quotient sheaf $\mathcal Q_x=\mathcal O_X/\mathcal I_x$ is supported at $x$, and its stalk is $\mathcal O_{X,x}/\mathcal I_{x,x}\cong\mathbb C$. Therefore
\begin{align*}
(M\otimes_{\mathcal O_X}\mathcal Q_x)_x
\cong M_{x,\mathrm{st}}\otimes_{\mathcal O_{X,x}}\mathbb C
\cong M|_x.
\end{align*}
Since $M\otimes_{\mathcal O_X}\mathcal Q_x$ is supported only at $x$, global sections are exactly the value of the stalk at $x$, so
\begin{align*}
H^0(X,M\otimes_{\mathcal O_X}\mathcal Q_x)\cong M|_x.
\end{align*}
This is why the quotient sheaf isolates the fibre.
Now apply the standard long exact sequence in sheaf cohomology to the short exact sequence
\begin{align*}
0 \longrightarrow M\otimes_{\mathcal O_X}\mathcal I_x \longrightarrow M \longrightarrow M\otimes_{\mathcal O_X}\mathcal Q_x \longrightarrow 0.
\end{align*}
The required hypothesis is that this is a short exact sequence of sheaves of abelian groups on $X$, and this holds by the standard exactness of tensoring with a locally free sheaf, since $M$ is a locally free $\mathcal O_X$-module of rank one. The long exact sequence gives the exact segment
\begin{align*}
H^0(X,M) \longrightarrow H^0(X,M\otimes_{\mathcal O_X}\mathcal Q_x) \xrightarrow{\delta_x} H^1(X,M\otimes_{\mathcal O_X}\mathcal I_x).
\end{align*}
The symbol $\delta_x$ denotes the connecting homomorphism attached to this short exact sequence.
Exactness at the middle term means
\begin{align*}
\operatorname{im}\left(H^0(X,M)\to H^0(X,M\otimes_{\mathcal O_X}\mathcal Q_x)\right)=\ker \delta_x.
\end{align*}
The vanishing hypothesis says
\begin{align*}
H^1(X,M\otimes_{\mathcal O_X}\mathcal I_x)=0.
\end{align*}
Therefore $\delta_x$ maps into the zero vector space, so every element of $H^0(X,M\otimes_{\mathcal O_X}\mathcal Q_x)$ lies in $\ker\delta_x$. Hence
\begin{align*}
\ker\delta_x=H^0(X,M\otimes_{\mathcal O_X}\mathcal Q_x).
\end{align*}
Combining this equality with exactness shows that
\begin{align*}
H^0(X,M)\longrightarrow H^0(X,M\otimes_{\mathcal O_X}\mathcal Q_x)
\end{align*}
is surjective. Under the identification
\begin{align*}
H^0(X,M\otimes_{\mathcal O_X}\mathcal Q_x)\cong M|_x,
\end{align*}
this is exactly the evaluation map $\operatorname{ev}_x:H^0(X,M)\to M|_x$. Thus every element of the fibre $M|_x$ is the value at $x$ of a global section of $M$.[/guided]