$\newcommand{\mf}[1]{\mathfrak{#1}}$
$\newcommand{\oO}{\mathcal{O}}$
$\newcommand{\ZZ}{\mathbb{Z}}$
$\newcommand{\Gal}{\text{Gal}}$
$\newcommand{\CC}{\mathbb{C}}$
$\newcommand{\QQ}{\mathbb{Q}}$
For (1), the definition goes like this: Suppose you have a galois extension of number fields $L/K$, then you have an extension of rings of integers $\oO_L/\oO_K$. The galois group $\Gal(L/K)$ acts as automorphisms on $L$ that fix $K$. Hence, it acts on $\oO_L$ and fixes $\oO_K$. Thus, for a prime $\mf{p}$ in $\oO_K$, and a prime $\mf{q}$ in $\oO_L$ containing $\mf{p}$, any $\sigma\in\Gal(L/K)$ must fix $\mf{p}$, and hence it must send $\mf{q}$ to another prime $\mf{q}'$ in $\oO_L$ which also contains $\mf{p}$ - ie, lies above $\mf{p}$. Thus, you may consider the subgroup of $\Gal(L/K)$ which fixes $\mf{q}$ - this is a subgroup of $\Gal(L/K)$, and is called the decomposition group $D_{\mf{q}/\mf{p}}$. In fact, $\Gal(L/K)$ acts transitively on the primes of $\oO_L$ lying above $\mf{p}$, and thus if $\sigma\in\Gal(L/K)$ sends $\mf{q}\mapsto\mf{q}'$, then $D_{\mf{q'}/\mf{p}} = \sigma(D_{\mf{q}/\mf{p}})\sigma^{-1}$, so the decomposition groups for primes lying above $\mf{p}$ are all conjugate (but still in general they are different subgroups).
Now, since $D_{\mf{q}/\mf{p}}$ fixes $\mf{q}$, it determines a well-defined action on $\oO_L/\mf{q}$, which is naturally seen to be a finite field extension of $\oO_K/\mf{p}$. In general, if you have a surjection of groups or rings $A\twoheadrightarrow B$, then an automorphism of $A$ descends to an automorphism of $B$ if and only if it fixes the kernel.
The elements of $D_{\mf{q}/\mf{p}}$ which induce the identity on $(\oO_L/\mf{q})/(\oO_K/\mf{p})$ are precisely the automorphisms $\sigma\in D_{\mf{q}/\mf{p}}$ such that $\sigma(x)\equiv x\mod \mf{q}$ for all $x\in\oO_L$. These automorphisms form a subgroup called the inertia group $I_{\mf{q}/\mf{p}}$.
Thus, there are inclusions $\{1\}\subset I_{\mf{q}/\mf{p}}\subset D_{\mf{q}/\mf{p}}\subset\Gal(L/K)$, which by galois theory corresponds to extensions of fields $L$ over $L^{I_{\mf{q}/\mf{p}}}$ over $ L^{D_{\mf{q}/\mf{p}}}$ over $K$. Going right to left, the prime $\mf{p}$ of $K$ decomposes completely in $L^{D_{\mf{q}/\mf{p}}}$, remains inert until $L^{I_{\mf{q}/\mf{p}}}$, and ramifies from there until $L$.
For (3), imagine the horizontal parabola given by, say, $x = y^2$, projecting onto the $x$-axis. At $x = 1$, you can solve the equation to find that $(1,1)$ and $(1,-1)$ are both points on the parabola which map to 1. Indeed, in general for $x = a$, the preimage of $a$ is $\{(a,\sqrt{a}),(a,-\sqrt{a})\}$. Thus, the preimage almost always has size 2, with one exception, namely when $a = 0$ (if $a < 0$ then the picture becomes deficient, but if you work over $\mathbb{C}$, you still have two (imaginary) solutions). Thus, this is an example of a finite map (ie, map with finite fibers, in this case generically of size 2, so you call the map a degree 2 map), which is ramified above $x = 0$, in the sense that above $x = 0$, there are fewer preimages than normal (in this case 1). In general this is what "ramification" means in geometry. An unramified map is one where the preimage (ie, fiber) above any point are all finite and have the same size.
What does this have to do with primes? Well, one of the great insights of Grothendieck is that prime ideals are kind of like points! For example, if you consider the ring $\CC[x]$, what are its prime ideals? Well, there is the zero ideal $(0)$, and then by the fundamental theorem of algebra, the other primes are all of the form $(x - a)$, where $a\in\CC$. Thus, we get a bijection between the maximal prime ideals of $\CC[x]$ and the points of $\CC$ by sending $(x-a)\mapsto a$. The $(0)$ ideal is kind of special - it's called the generic point, and in a sense represents the entirety of $\CC$.
Note that the rings $\CC[x]$, $\oO_K$ are very similar! They're both Dedekind domains, and in fact $\CC[x]$ is even a PID (class number 1). The fraction field of $\CC[x]$ is $\CC(x)$ and is analogous to $K$. A galois extension of $\CC(x)$ looks like $\CC(x)[y]/f(y)$ and is analogous to the extension $L/K$. Then, $\oO_L$ is analogous to the integral closure of $\CC[x]$ in $\CC(x)[y]/f(y)$.
If you take our example with the parabola, you could set $f(y) = y^2 - x$, then $L' := \CC(x)[y]/(y^2-x)$ is a quadratic extension of $K' := \CC(x)$. It is galois, with galois group of order 2 (hence isomorphic to $\mathbb{Z}/2\mathbb{Z}$, and is generated by the automorphism which fixes $\CC(x)$ and sends $y\mapsto -y$. The integral closure of $\oO_{K'} := \CC[x]$ inside $L'$ in this case is just $\oO_{L'} := \CC[x,y]/(y^2-x)$.
For the prime $\mf{p}_a := (x-a)$ of $\CC[x]$, then there are two primes of $\oO_{L'}$ lying above $\mf{p}_a$, namely $(y-\sqrt{a}),(y+\sqrt{a})$, which are different primes, unless $a = 0$, in which case they are the same.
Exercise: Compute the inertia and decomposition groups for primes of $\oO_{L'}$ lying above $(x - 1),(x-2)$ and $(x-0) = (x)$. You'll find that in the first two cases the decomposition group and inertia group are both trivial. In the second case, the decomposition group and inertia group are both the whole galois group. Note that in this situation primes never remain inert, simply because all the residue fields of $\CC[x]$ are just $\CC$, which is algebraically closed. Note that in each case the galois group acts transitively on the primes of $\oO_{L'}$ lying above a prime of $\oO_{K'}$.
Exercise: Play the same game above with $\CC[x]$ replaced by $\QQ[x]$. Note that $\QQ[x]$ is still a Dedekind domain (in fact a PID), so all the definitions of decomposition/inertia still make sense. You should find that $(x-1)$ splits, $(x-2)$ remains inert, and $(x-0) = (x)$ is ramified.
Remark: In the language of schemes, the topological space $\CC$, viewed as a scheme, is Spec $\CC[x]$, where Spec $R$ for a ring $R$ is a topological space whose underlying set is the set of prime ideals of $R$. The topology however is called the Zariski topology and in the case of Spec $\CC[x]$, does not agree with the complex topology on $\CC$. However, the cool thing here is that the traditionally geometric notions of dimension, smoothness, connectedness, compactness, fundamental groups, picard groups, tangent spaces, differential forms,...etc can all be defined purely algebraically for RINGS. From this viewpoint, the ring of integers of an algebraic number field is a smooth connected non-compact curve. $\ZZ$ has trivial fundamental group, since all rings of integers are ramified over $\ZZ$ (this follows I believe from the fact that the discriminant of a number field is never a unit, and comes from the geometry of numbers (NOT algebraic geometry)).
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
Let $\mathfrak{q}$ be a prime of $\mathcal{O}_L$ lying over $\mathfrak{p}$, a prime in $\mathcal{O}_K$.
Let $E$ be the inertial group, i.e. $$ E = \{\sigma \in G : \sigma (\alpha) \equiv \alpha \mod \mathfrak{q} \}.$$
Let $D$ be the decomposition group, i.e. $$ D = \{\sigma \in G : \sigma(\mathfrak{q}) = \mathfrak{q} \}. $$
Moreover, let $L_E$ and $L_D$ denote the fixed fields of $E$ and $D$ respectively, and let $\mathfrak{q}_E = \mathfrak{q} \cap L_E$ and define $\mathfrak{q}_D$ in a similar manner. It'll turn out that you have the following: \begin{align*} (L:L_E) &= e\\ (L_E:L_D) &= f\\ (L_D:K) &= r \end{align*} where $r$ is the number of primes that lie over $\mathfrak{p}$.
Moreover, you'll have that $\mathfrak{q}$ is totally ramified over $\mathfrak{q}_E$, i.e. $e( \mathfrak{q} | \mathfrak{q}_E ) = e$ and $f(\mathfrak{q} | \mathfrak{q}_E ) = 1$.
We also have $e( \mathfrak{q}_E | \mathfrak{q}_F ) = 1$ and $f(\mathfrak{q}_E | \mathfrak{q}_F ) = f$.
Finally, we have $e( \mathfrak{q}_F | \mathfrak{p} ) = 1$ and $f(\mathfrak{q}_F | \mathfrak{p} ) = 1$.
Now, let $K'$ be an arbitrary intermediate field, i.e. $K \leq K' \leq L$ and let $\mathfrak{p}' = \mathfrak{q} \cap K'$. The following is taken from Daniel Marcus' Number Fields: we then have
$L_D$ is the largest intermediate field $K'$ s.t. $e(\mathfrak{p}' | \mathfrak{p}) = f(\mathfrak{p}' | \mathfrak{p}) = 1$.
$L_D$ is the smallest $K'$ s.t. $\mathfrak{q}$ is the only prime lying over $\mathfrak{p}'$.
$L_E$ is the largest $K'$ s.t. $e(\mathfrak{p}' | \mathfrak{p}) = 1$.
$L_E$ is the smallest $K'$ s.t. $\mathfrak{q}$ is totally ramified over $\mathfrak{p}'$.
In summary, $E$ and $D$ allow us to decompose the splitting into subfields where each step ($r,e,$ and $f$) occurs in a separate place