The definition that sheaf of $\mathcal{O}_X$ module is finite type is in here.
Understand a result about sheaf of modules of finite type
I'm also trying to understand this.
Let $X$ be a topological space. Let $0\to\mathcal{F}\xrightarrow{k_1}\mathcal{G}\xrightarrow{k_2}\mathcal{H}\to 0$ be a short exact sequence of $\mathcal{O}_X$-modules. If $\mathcal{F}$ and $\mathcal{H}$ are finite type, so is $\mathcal{G}$.
but I don't understand the description in stacks, so I'm now reading a proof of another reference.I'll write down the proof.
Take $x\in X$, there are open neighborhoods $x\in U',U''$ and $m,n>0$ such that
\begin{align}
{\mathcal{O}_X}^m|_{U'}\xrightarrow{\alpha}\mathcal{F}|_{U'}\to 0,
{\mathcal{O}_X}^n|_{U'}\xrightarrow{\beta}\mathcal{H}|_{U'}\to 0
\end{align}
are exact. We may assume $U''=U'$(considering taking stalk).
Let $e_1\dots e_n\in \mathcal{O}_X(U')^n$ is a standard basis,then there is an open neighborhood $x\in U\subset U'$ and $f_1\dots f_n\in\mathcal{G}(U)$ such that $\beta(U)(e_i|_U)={k_2}(U)(f_i)$ for $i=1\dots n$.Now we define $\gamma\colon {\mathcal{O}_X}^n|_{U}\to\mathcal{G}|_{U}$ as $\gamma (V)(e_i|_V)\colon= f_i|_V$ for open set $V\subset U$.
Then ${k_1}|_U\circ\alpha|_U+\gamma$ gives ${\mathcal{O}_X}^m|_{U}\oplus{\mathcal{O}_X}^n|_{U}\to\mathcal{G}|_{U}$ and this is surjective$\dots (\ast)$.
I don't understand $(\ast)$.
Best Answer
All that's going on here is that we're using the fact that if $0\to A\to B \to C\to 0$ is an exact sequence of $R$-modules where $A,C$ are finitely generated, then $B$ must be finitely generated as well.
Proof: If $A,C$ are finitely generated, then pick surjections $s_A:R^m\to A$ and $s_C:R^n\to C$. Now consider the following commutative diagram:
$$\begin{array}{ccccccccc} & & R^m & & & & R^n & & \\\ & & \downarrow & & & & \downarrow & & \\\ 0 & \xrightarrow{} & A & \xrightarrow{f} & B & \xrightarrow{g} & C & \xrightarrow{} & 0 \end{array}$$
where the vertical arrows are the surjections. Then as $B\to C$ is a surjection and $R^n$ is projective, we get map $h:R^n\to B$ so that the composition $g\circ h$ is the same as the surjection $s_C:R^n\to C$. We can then define a map from $s_B:R^{m+n}=R^m\oplus R^n$ to $B$ as follows: send $(x,y)\mapsto (f\circ s_A)(x) + h(y)$.
To show that this is surjective, consider an element $b\in B$. Then there's some $y\in R^n$ so that $s_C(y)=g(b)$ by definition of surjectivity of $s_C$. Now consider $b-h(y)$: this element must map to $0$ in $C$ under $g$ by linearity: $g(b-h(y))=g(b)-g(h(y))=g(b)-s_C(y)=g(b)-g(b)=0$. So $b-h(y)\in \ker(g)$, and thus by exactness $b-h(y)$ must be in the image of $f$, say $f(a)=b-h(y)$. By surjectivity of $s_A$, there is some $x\in R^m$ so that $s_A(x)=a$. This construction exactly gives $s_B(x,y)=b$, as $s_B(x,y)-b=f(a)+h(y)-b=f(a)-f(a)=0$. $\blacksquare$
This is the same proof (with different notation) as is going on in the quoted portion of your post.