Let $j:\mathbb{C}^* \hookrightarrow \mathbb{C}$ be the standard open embedding. If we consider the perverse sheaf $F=\mathbb{Q}_{\mathbb{C}^*}[1]$. It is possible to show that the higher direct image $Rj_*F$ and the higher direct image with compact support $Rj_{!}F $ are both perverse sheaves.
(I'm referring to the following notes by de Cataldo page 26 for the notations http://www.math.stonybrook.edu/~mde/MyPublishedPapers/PCMI_2015_Lecture_Notes.pdf).
We have then a natural map $f:Rj_!F \to Rj_*F$ in the category $Perv(\mathbb{Q}_{\mathbb{C}})$: I'd like to compute its kernel and cokernel. My idea would have been the following:
Write down a short exact sequence: $ 0 \to Rj_!F \to Rj_*F \to G \to 0$ for some complex of sheaves $G$.
From this we get a distinguished triangle $Rj_!F \to Rj_*F \to G \to $ and in this situation we have an explicit way to determine $ker f $ and $coker f$.
However, I was not able to find such a short exact sequence. I was also wondering whether there is a simpler and more general approach to compute kernel and so on of maps between perverse sheaves.
I'm afraid I don't know of a simpler and more general approach, but I can do this case.$ \newcommand{\c}[1]{\mathbb{Q}_{#1}} \newcommand{\C}{\mathbb C} \DeclareMathOperator{\coker}{coker} \newcommand{\Id}{\mathrm{Id}} \newcommand{\H}{{}^p\mathcal{H}} $ A useful thing to try is to use the attaching triangles (page 21 of the notes you're using). Let $i$ be the inclusion of $0$ in $\C$. Let me just call $Rj_!j^*\to \Id \to i_*i^*$ "the first" and $i_*i^!\to \Id \to Rj_*j^*$ "the second" attaching triangle. Also please assume that all functors are derived (I'm not sure if I'm supposed to write $i^*$ or $Li^*$...).
Applying the first triangle to $Rj_*F$, we have the distinguished triangle: $$ Rj_!j^*Rj_*F\to Rj_*F \to i_*i^*Rj_*F. $$ Since $j^*\circ Rj_*\cong \Id$, this triangle is the one you are talking about in your question, so by the uniqueness of the cones, it follows that there is an isomorphism $i_*i^*Rj_*F\cong G$.
We can find $i^*Rj_*F$ using the definitions. By definition of pullback: $$ i^*Rj_*F = \varinjlim_{U\ni 0} R\Gamma^\bullet(U;Rj_*F), $$ where the limit is taken over the open neighborhoods of $0$ (so we can take $U$ to be disks). Then, if we call $a$ the constant map to a point, $$ R\Gamma^\bullet(U;Rj_*F) \cong Ra_*(Rj_*F) \cong R(a\circ j)_*F \cong R\Gamma^\bullet(\C^*;F) \cong H^\bullet(\C^*;\mathbb Q)[1]\cong \mathbb Q[1]\oplus \mathbb Q[0]. $$ This is a complex with one-dimensional cohomologies in degrees $-1$ and $0$ (due to the shift) and zero elsewhere. The limit $i^*Rj_*F$ can be taken over disks, so it is a limit of a system of isomorphisms, and the limit itself is isomorphic to the complex above as well.
This shows that $i^*Rj_*F$ is quasiisomorphic to $\mathbb Q[1]\oplus \mathbb Q[0]$, so $G$ is the pushforward of this sheaf, i.e. a sum of two 1-dimensional skyscraper sheaves, in degrees $-1$ and $0$.
Since a skyscraper sheaf in degree $0$ is perverse in this case, taking the long exact sequence in perverse cohomology of your distinguished triangle yields the following exact sequence in the category of perverse sheaves: $$ 0\to i_*\mathbb Q \to Rj_!F \to Rj_*F \to i_*\mathbb Q\to 0. $$ So the kernel and cokernel are both one-dimensional skyscraper sheaves.
Some ways to approach this which might be more general (I'm not sure what you want to generalize to) could be to use a de Rham resolution to write down $Rj_*F$ explicitly and finding $G$ this way, or to use the Riemann-Hilbert correspondence, in which case perverse sheaves would turn into D-modules (not just complexes of D-modules), so you can find the kernel and cokernel without any triangulated business.