I have two different definition of $\mathcal{O}_X$-module of finite type.
One definition is that given $\mathcal{F}$ an $\mathcal{O}_X$-module we have that $\forall x \in X$ there exists $U$ open neighborhood s.t. $$ \mathcal{O}_U^{(I)} \rightarrow \mathcal{F}|_U \rightarrow 0 $$ is exact and $I$ is finite.
The other one is that $\forall x \in X$ there exists $U$ open neighborhood s.t. $\mathcal{F}|_U$ is generated by finitely many sections.
However I don't know how to really prove the equivalence of these definitions. My idea was looking at the fact that the image of the basis of $\mathcal{O}_U^{(I)}$ is a generating set.
The answer lies in the definition of "generated by finitely many sections". As I understand, $\mathcal{F}|_U$ is generated by finitely many sections if, there are sections $s_1,\dots, s_n$ over $U$ such that $\mathcal{F}|_U$ is the smallest sheaf of $\mathcal{O}_U$-modules that contains all the $s_i$. In other words, if $\mathcal{G}$ is a subsheaf of $\mathcal{O}_U$-modules of $\mathcal{F}|_U$ that contains all the $s_i$, then $\mathcal{G}=\mathcal{F}|_U$.
For each $s_i$, there is a morphism of sheaves of $\mathcal{O}_U$-modules $\varphi_i:\mathcal{O}_U\to \mathcal{F}|_U$, defined by $f\mapsto fs_i$ (here $f$ is a local section of $\mathcal{O}_U$). Taking coproduct, we have $$\varphi:\bigoplus_{i=1}^n \mathcal{O}_U \to \mathcal{F}|_U$$ It is easy to see that $\operatorname{Im}\varphi$ is a subsheaf of $\mathcal{F}|_U$ that contains all the $s_i$, so $\operatorname{Im}\varphi=\mathcal{F}|_U$ and $\varphi$ is surjective, which is indeed the other definition.
You may be more comfortable with this definition. We say a sheaf of $\mathcal{O}_U$-modules $\mathcal{F}$ is generated by sections $s_i,i\in I$ if for every $x\in U$, $\mathcal{F}_x$ is generated by $(s_i)_x$ as a $\mathcal{O}_{U,x}$-module. Under this definition, the morphism $\varphi$ constructed above is surjective on every stalk, hence is surjective.
For more information, I suggest you reading this Stack Project entry.