I think you have yourself answered the most of the questions. But, for the sake of completeness, I would like to write an answer. When $p$ cannot divide $ [l:k]_{insep}$, there could be no inseparable extensions between $l$ and $k$, hence $l/k$ is separable, showing that $(i)$ and $(ii)$ are equivalent. Other implications have already been explicitly answered by you.
P.S. One could show that $l/k$ is separable by the fact that inseparable extensions occur only when the characteristic is $>0$, and when the degree is divisible by the prime characteristic. Hence the conclusion.
Notice that this degree needs not be a power of the prime, as indicated by QiL'8.
We are looking for a collection of ramification and splitting behaviours over a single prime $P$, all of which are possible, but rarely combine in a single easily-calculated example. I have produced a case where each of the things you ask for does indeed occur, in order to illustrate the general case. But I should point out that it is easy to produce individual lower-degree examples demonstrating separately each of the requirements you have placed on the primes above $P$.
Recall the well-known formula describing the splitting of a prime $P$ in an extension $L/K$ in terms of its ramification and residue field extension indices:
\begin{equation*}
\Sigma_{Q\mid P}\ e_{Q}f_{Q} = [L:K].
\end{equation*}
As you have pointed out, when the extension is Galois all of the $e_Q$ are equal to some fixed $e$, and all of the $f_Q$ are equal to a fixed value $f$; so this reduces to
\begin{equation}
efg = [L:K].
\end{equation}
where we have used the standard notation $g$ for the number of primes of $L$ above the prime $P$ of $K$ in a Galois extension.
We would like to see different types of behaviour at each of the levels $L$, $L_D$ and $L_E$ for some prime $Q$ of $L$ above $P$, which implies that we need $e\geq2$ and $f\geq2$. Moreover in order to have any chance of seeing different outcomes above the same prime $P$, we need that $P$ split into at least 2 distinct primes at the level $L_D$, otherwise $L_D=K$. Hence we also need $g\geq2$. So the degree of the extension $L/K$ needs to be at least 8.
Finally, as you also point out, we need that the Galois group Gal($L/K$) contain non-normal subgroups in order that any prime have a chance that the fixed field of its decomposition and/or inertia groups be non-Galois; otherwise we are just in a lower-degree version of the above formula. It is not sufficient, by the way, that the extension simply be non-Abelian, since for example the quaternion group Q8 is non-Abelian but has no non-trivial non-normal subgroups.
All of this forces the extension to have degree at least 16 (the non-normal subgroups of the dihedral group D4 of order 8 do not allow enough varied behaviour). We also would like an extension whose Galois group is furnished with lots of non-normal subgroups. So the easiest place to start would be something with Galois group S4. In order to simplify things let us assume that $K=\mathbb{Q}$, and take some "general" quartic extension, the simplest interesting one of which might be the splitting field $L$ of the polynomial $q(x) = x^4+x+1$.
I used MAGMA for the following calculations. This extension $L/\mathbb{Q}$ is ramified only above $P=229$, splitting into six primes with residue field extension (="inertia") degree $f=2$ and ramification degree $e=2$. Choosing any of these primes $Q$ say we have an inertia group $E(Q\mid P)$ which is cyclic of order 2 ($\cong C_2$) and a decomposition group $D(Q\mid P)$ which is isomorphic to $C_2^2$.
We then calculate that there is always a prime $Q_D'$ of $L_D$ over $P$ with $e_{Q_D'}=f_{Q_D'}=2$.
Similarly there is always a prime $Q_E'$ of $L_E$ over $P$ with $e_{Q_E'}=2$ and $f_{Q_E'}=1$.
Best Answer
One useful fact is that if $L/K$ is a totally ramified extension of local fields, then $\mathcal{O}_L \simeq \mathcal{O}_K[\pi_L]$ for a root $\pi_L$ of some Eisenstein polynomial over $K$, with $\pi_L$ a uniformizer for $L$. Hence the residue field $\mathcal{l}$ of $L$ is given by $$\mathcal{l} = \mathcal{O}_{L}/(\pi_L) \simeq \mathcal{O}_K[\pi_L]/(\pi_L) \simeq \mathcal{O}_K/((\pi_L) \cap \mathcal{O}_K) = \mathcal{O}_{K}/(\pi_K) = k$$ where $\pi_K$ is a uniformizer for $K$, so we have a natural identification of the residue field of $L$ and the residue field of $K$.
In the diagram, $L_{\mathfrak{P}}$ is a totally ramified extension of $L_{\mathfrak{P}}^{I(\mathfrak{P})}$ since the Galois group is the inertia group $I(\mathfrak{P})$. Hence we may identify the residue fields of these two local fields.