Notation : if $A \subset \mathbb R^2$ then $A^* := A \backslash \{0\}$.
Let $L$ a local system on $D$ with stalk $V$, and $j : D^* \to D$ the inclusion.
I want to understand the difference between $j_* L$ and $j_! L$, more precisely the image of the natural morphism $j_!L \to j_*L$.
$j_*L$ is simply the sheaf $ W \mapsto L(W^*)$. $j_!L$ is the subsheaf of $j_*$ with compact support in $V$. But since sections are locally constant, shouldn't we have that the only compact support-section are the trivial one ? I think I probably miss something obvious or I don't really understand what the section of a local system are. Thanks in advance for any help !!
The definition of $j_!L$ is not exactly what you said. There is several definitions, but I guess the one you had in mind is the following : $j_!L$ is the subsheaf of $j_*L$ such that $$j_!L(W)=\{s\in j_*L(W)=L(W^*) \text{ such that } \operatorname{Supp}(s)\rightarrow W \text{ is proper }\},$$ where $\operatorname{Supp}(s)\subset W^*$ is the support of $s$ seen as section in $L(W^*)$.
Note in particular that if $W$ does not contain $0$, then, a section of $j_*L$ would automatically have proper support on $W$. For example, let $L=\mathbb{Z}$ be the constant sheaf on $D^*$. Then $j_*\mathbb{Z}(D^*)=\mathbb{Z}(D^*)=\mathbb{Z}$ and the constant function 1 has $D^*$ as support which is obviously proper on $D^*$.
Hence $j_!L(W)=j_*L(W)$ if $0\not\in W$.
However, if $W$ does contain 0, then the section of $j_!L(W)$ needs to have a support disjoint from 0, otherwise, the support will not be closed in $D$. So if $L$ is a local system and $W$ is connected, there is no section at all.
In other words, a section $s\in j_*L(W)$ is in $j_!L(W)$ if $s_0=0$. We find here another definition of $j_!L$ : it is the kernel of $j_*L\rightarrow i_*i^{-1}j_*L$ where $i:\{0\}\rightarrow D$ is the inclusion.