The applications are HUGE! Let me just mention two here.
1) Chebotarev density answers an important question in the study of number fields:
What information is contained within the splitting primes?
To make this slightly more precise, let us define for an extension $L/K$ of number fields the set of splitting primes as follows:
$$\text{Spl}(L/K)=\left\{\mathfrak{p}\text{ a prime of }\mathcal{O}_K:\mathfrak{p}\text{ splits completely in }\mathcal{O}_L\right\}$$
This is an object of extreme importance to us in number theory. In fact, one might say that given an extension $L/K$ our main number theoretic interest in $L$ is to determine $\text{Spl}(L/K)$. So, an obvious question presents itself: what information precisely is contained in the set $\text{Spl}(L/K)$?
The answer is beautiful:
Theorem: Let $L_1,L_2/K$ be be two number field extensions of $K$. Then, the following are equivalent:
- $L_1$ and $L_2$ have the same Galois closures.
- The sets $\text{Spl}(L_1/K)$ and $\text{Spl}(L_2/K)$ are equal.
- The sets $\text{Spl}(L_1/K)$ and $\text{Spl}(L_2/K)$ are almost equal.
Here 'almost equal' means that there are only finitely many primes not contained in either.
So, this theorem is INCREDIBLE! It tells you that the number theoretic question we have always been interested, which primes split completely, is not just a question of number theoretic significance, but of field theoretic significance. In particular, to know a field's Galois closure (field theory) is the same thing as knowing its set of split primes (number theory).
This also shows that while, a priori, knowing just the split primes only tells you some number theoretic information it doesn't tell you all. Namely, the splits primes shouldn't, a priori, tell you about ramified primes, etc. But, if your extension is Galois, then the above tells you that the split primes know about $L$ itself, and so, of course, know about the other number theoretic data.
OK, excellent, this is a beautiful theorem. What does it have to do with Chebotarev density? Well—everything! Namely, the proof of this theorem is essentially Chebotarev density. Let me give a sketch below:
Proof: Suppose first that $L_1$ and $L_2$ have the same Galois closure, call it $L$. Then, elementary algebraic number theory shows that
$$\text{Spl}(L_1/K)=\text{Spl}(L/K)=\text{Spl}(L_2/K)$$
which shows that 1. implies 2.
Conversely, suppose that $\text{Spl}(L_1/K)=\text{Spl}(L_2/K)$. Then, if $L_i'$ denotes the Galois closures of $L_i$, then $\text{Spl}(L_1'/K)=\text{Spl}(L_2'/K)$.
But, note then that, again by basic number theory, this implies that
$$\text{Spl}(L_1'/K)=\text{Spl}(L_1'L_2'/K)=\text{Spl}(L_2'/K)$$
But, by considering Chebotarev density, since all of these extensions are Galois, we deduce that the following densities are equal
$$\frac{1}{[L_1':K]}=\frac{1}{[L_1'L_2':K]}=\frac{1}{[L_2':K]}$$
which, in particular, shows that $[L_i':K]=[L_1'L_2':K]$ which implies that $L_1'=L_1'L_2'=L_2'$ which shows that $L_1'=L_2'$ as desired. $\blacksquare$
As an application of this theorem, we deduce the main idea of Class Field Theory:
Idea: Extensions of $K$ for which data about the extension is entirely 'internal to K' are precisely the abelian extensions.
A rigorous example of this is:
Theorem: If $K/\mathbb{Q}$ is a Galois extension such that the splitting behavior is determined $\mod N$ for some integer $N$, then $K$ is abelian (in fact contained in $\mathbb{Q}(\zeta_N)$).
If you want to read more about this application, you can see my blog post here.
2) The second idea comes from the theory of Galois representations. Namely, let us say that an $\ell$-adic Galois representation of $K$ (a number field) is a continuous group homomorphism
$$\rho:G_K\to\text{GL}_n(\overline{\mathbb{Q}_\ell})$$
These are EXTREMELY important in modern number theory, in ways in which I won't go into here. But, an interesting question is how little information is needed to determine $\rho$. What data do we need to compute to know that we've uniquely characterized $\rho$?
If $\rho$ is unramified almost everywhere, the answer is very satisfying. We say that $\rho$ is unramified almost everywhere if it factors through $G_{K,S}=\text{Gal}(K^S/K)$, where $K^S$ is the maximal extension of $K$ unramified outside of $S$, for some finite set $S$ of primes of $K$.
These are the most important types of Galois representations, one generally only considers representations of this form. In particular all 'geometric Galois representations', those coming from geometry (e.g. the Tate module of an abelian variety), are unramified almost everywhere.
The result is then the following:
Theorem: Let $\rho_1:G_{K,S}\to\text{GL}_n(\overline{\mathbb{Q}_\ell})$ and $\rho_2:G_{K,T}\to\text{GL}_n(\overline{\mathbb{Q}_\ell})$be two unramified almost everywhere Galois representations. Then, $\rho_1=\rho_2$ if and only if
$$\text{tr}(\rho_1(\text{Frob}_\mathfrak{p}))=\text{tr}(\rho_2(\text{Frob}_\mathfrak{p}))$$
for all $\mathfrak{p}\notin S\cup T$.
Let me just explain the above notation. For $\mathfrak{p}\notin S$ there is a well-defined Frobenius conjugacy class $\text{Frob}_\mathfrak{p}\in G_{K,S}$. Indeed, one can understand this in elementary terms as writing $K^S$ as a union of finite extensions of $K$. Then, since each of these extensions are unramified at $\mathfrak{p}$ they have a Frobenius conjugacy class, and so we obtain one in the union.
Then, $\text{tr}(\rho(\text{Frob}_\mathfrak{p}))$ denotes the trace of the image of any element of $\text{Frob}_\mathfrak{p}\subseteq G_{K,S}$. It, of course, is independent of choice since the trace function ignores conjugation.
Thus, this theorem tells us that the huge amount of data encompassed in $\rho$ is, in fact, contained in this MUCH smaller set of the traces of the Frobenii. Amazing!
The proof relies on two facts:
a) The Brauer-Nesbitt theorem.
b) The fact that the Frobenius conjugacy classes $\{\text{Frob}_\mathfrak{p}\}_{\mathfrak{p}\notin S}$ are dense in $G_{K,S}$.
The first of these is just a classic result in algebra. But, b) is, for all intents and purposes the 'same thing' as Chebotarev density. Exercise: use Chebotarev density to prove b)!
Best Answer
As you note, the assumption that $f(x)$ splits into linear factors modulo almost all $\mathfrak p$ implies that the decomposition group at $\mathfrak P$ is trivial for almost all $\mathfrak P$. (This is standard algebraic number theory, and presumably Milne expects the reader to realize this without any explanation being necessary.)
With this fact in hand, Milne's claim follows directly from Cebotarev:
Let $G$ be the splitting field of $L$ over $K$. Cebotarev says that any element of $G$ can be realized as a Frobenius element (in particular as an element of the decomposition group $G({\mathfrak P})$) for infintely many $\mathfrak P$. But since $G(\mathfrak P)$ is trivial for all but finitely many $\mathfrak P$, we see that every element of $G$ must be trivial, hence that $G$ is trivial, i.e. that $L = K$, i.e. that $f(x)$ splits over $K$.
(If this is the argument you had in mind, then I'm not sure what you're asking.)
A side remark:
I'm not sure what you mean when you write that the factorization of $f(X)$ does not necessarily give the factorization of a prime $\mathfrak p$. As I mentioned in my answer to this question of yours, if $f(x)$ is any element of $K[x]$, then we may write $f(x) \in \mathcal O_K[1/a][x]$, for some $a \in \mathcal O_K\setminus \{0\}.$ Choosing $a$ appropriately, we may furthermore assume that the discrminant of $f$ is invertible in $\mathcal O_K[1/a]$, so that for every prime $\mathfrak p$ of $\mathcal O_K[1/a]$, the reduction of $f(x)$ modulo $\mathfrak p$ is separable.
The factorization of any prime $\mathfrak p$ of $\mathcal O_K[1/a]$ in the splitting field $L$ of $f(x)$ then corresponds precisely to the factorization of $f(x)$ modulo $\mathfrak p$.
Of course, the primes of $\mathcal O_K[1/a]$ are precisely the primes of $\mathcal O_K$ which do not divide $a$, and so consist of all but finitely many prime of $\mathcal O_K$. Thus if one is willing to throw away finitely many primes (which is obviously harmless in the case of e.g. Milne's claim), then one can just work with the ring $\mathcal O_K[1/a]$ and apply Kummer's lemma in this ring. Everything proceeds just as if $f(x)$ has integral coefficients (and there is no need whatsoever to worry about generators of the ring of integers in either $K$ or $L$; e.g., since the discriminant of $f(x)$ is invertible in $\mathcal O_K[1/a]$, the ring $\mathcal O_L[1/a]$ is generated by the roots of $f(X)$ over $\mathcal O_K$).