Welcome new contributor. It is best to think about these kinds of examples on one's own. Thus, *please try this for yourself before reading the following*.

Let $(Y,Z)$ be any pair of a Noetherian scheme $Y$ and a nonempty closed subset $Z$ that is nowhere dense. For instance, let $Y$ be the following Spec of a DVR, $Y=\text{Spec}\ k[t]_{\langle t\rangle}$, and let $Z$ be the singleton set $\{z\}$ of the $k$-point of the unique maximal ideal. Denote by $i$ the inclusion continuous function from $Z$ to $Y$.

Consider the sheaf of $\mathcal{O}_{Y}$-modules $i_*i^{-1}\mathcal{O}_{Y}$. For every open subset $U$ of $Y$ that does not intersect $Z$, the only section of this sheaf on $U$ is the zero section, and thus the stalk at every $y\in U$ is the zero module. Also, for every $z\in Z$, the stalk at $z$ equals $\mathcal{O}_{Y,z}$.

Now form the sheaf of $\mathcal{O}_Y$-modules $\mathcal{O}_X:=\mathcal{O}_Y \oplus \left( i_*i^{-1}\mathcal{O}_Y\cdot \epsilon \right)$, where $\epsilon$ is just a placeholder. Give this the unique structure of $\mathcal{O}_Y$-algebra such that $\epsilon\cdot \epsilon$ equals $0$ and such that the following natural inclusion map is a morphism of $\mathcal{O}_Y$-algebras, $$\mathcal{O}_Y \hookrightarrow \mathcal{O}_Y \oplus \left( i_*i^{-1}\mathcal{O}_Y\cdot \epsilon \right), \ \ f\mapsto (f,0\cdot \epsilon).$$

For every $y\in Y\setminus Z$, the stalk $\mathcal{O}_{X,y}$ equals $\mathcal{O}_{Y,y}$ as an $\mathcal{O}_{Y,y}$-algebra. Also, for every $z\in Z$, the $\mathcal{O}_{Y,z}$-algebra $\mathcal{O}_{X,z}$ equals $\mathcal{O}_{Y,z}\oplus \mathcal{O}_{Y,z}\cdot \epsilon$ with $\epsilon\cdot \epsilon = 0$. In every case, the stalk is a Noetherian local ring.

In the special case that $Y$ equals $\text{Spec}\ k[t]_{\langle t\rangle}$ and $Z$ equals the singleton of the closed point, then also $\mathcal{O}_X(Y)$ equals $k[t]_{\langle t \rangle} \oplus k[t]_{\langle t \rangle}\cdot \epsilon$, which is Noetherian. Also $\mathcal{O}_X(Y\setminus Z)$ equals $k(t)$, which is Noetherian. Thus, also the ring of sections is Noetherian for every open subset.

There is a natural retraction of the above algebra homomorphism, $$\mathcal{O}_X\to \mathcal{O}_Y, \ \ (f,g\cdot \epsilon) \mapsto f.$$ Consider the $\mathcal{O}_X$-module morphism from $\mathcal{O}_X$ to itself that multiplies by the global section $(0,1\cdot \epsilon)$. Denote the kernel sheaf by $\mathcal{K}$. Denote by $\mathcal{K}'$ the associated $\mathcal{O}_Y$-module $\mathcal{K}\otimes_{\mathcal{O}_X}\mathcal{O}_Y$.

If $\mathcal{K}$ is locally finitely generated as an $\mathcal{O}_X$-module, then also $\mathcal{K}'$ is locally finitely generated as an $\mathcal{O}_Y$-module. However, $\mathcal{K}'$ is the extension by zero of the structure sheaf on $Y\setminus Z$. This is not a finitely generated $\mathcal{O}_X$-module. In fact, for every irreducible open neighborhood $U$ that intersects $Z$, the sections of $\mathcal{K}'$ on $U$ are zero, yet they are nonzero for $U\setminus Z$. Thus, every map $\mathcal{O}_U^{\oplus n}\to \mathcal{K}'|_U$ is the zero map, yet the restriction of $\mathcal{K}'|_U$ to $U\setminus Z$ is nonzero.