How can the commutator operation not be transitive? I noticed the following:
$$[L_{+},L^2]=0,\qquad [L_{+},L_3]\neq 0,\qquad  [L^2,L_3]=0.$$
This would suggest, that $L^2,L_+$ have a common system of eigenfunctions, and so do $L^2,L_3$, but $L_+,L_3$ don't. How is that possible?
 A: Commutativity is not a transitive relation: If operator $A$ commutes with $B$ and $C$, 
$$AB=BA \quad\text{and}\quad AC=CA,$$ 
then there is no reason that $B$ and $C$ should commute. 
Example: Take $A=y$, $B=x$, and $C=p_x$.
In particular, if commuting selfadjoint operators $A$ and $B$ have a common basis of orthonormal eigenvectors, and if commuting selfadjoint operators $A$ and $C$ have a common basis of orthonormal eigenvectors, then these two bases need not be the same if the spectrum of $A$ is degenerate.
A: NOTE. Since $L_+$ is not normal (normal means $A^\dagger A = AA^\dagger$) it does not admit a basis of orthonormal eigenvectors. However your question can be safely restated replacing $L_+$ for $L_2$ and I will assume it henceforth.
The most elementary case of this phenomenon is given by a triple of normal matrices  in $\mathbb C^n$:
$$cI,A,B$$
with $[A,B]\neq 0$ and where  $c\in \mathbb C$ is an arbitrarily fixed number. $A$ has a common basis of eigenvectors with $cI$: Every basis of eigenvectors of $A$ is such basis. Similarly, every basis of eigenvectors of $B$ is also a basis of eigenvectors of $cI$. However, though it could happen for some vector, there cannot exist a whole basis of eigenvectors in common with $A$ and $B$, otherwise referring to that basis $A$ and $B$ would be in diagonal form and thus $[A,B]=0$, which is forbidden by hypotheses. 
All that is possible thanks to the fact that the eigenspaces of $cI$ are (maximally) degenerate. Two vectors $u$ and $v$ with the same eigenvalue ($c$) of respect to $cI$ remain eigenvectors of $cI$ with the same eigenvalue even if linearly composed: $au+bv$. Nevertheless if $u$ and $v$ are eigenvectors of $A$, in general $au+bu$ is not, but it could be an eigenvector of $B$ (remaining, as said, an eigenvector of $cI$)
The situation is essentially the same when dealing with $L^2$ and $L_2,L_3$.
The eigenspaces $\cal H_l$ of $L^2$ are  degenerate and, in each eigenspace, $L^2$ is represented by $l(l+1)I$. 
Moreover, as $[L^2,L_i]=0$, each eigenspace $\cal H_l$ is invariant under the action of $L_i$. I mean $L_i({\cal H}_l)\subset \cal H_l$.
Restricting to $\cal H_l$, we find the situation I outlined above: $L$ is represented by $cI$ and $L_2, L_3$ are represented by non commuting  operators $A$ and $B$.
