Let $R$ be a Dedekind Ring and $K = \operatorname{Quot}(R)$. Let $\mathcal{I}_K$ be the ideal group and $C \ell_K$ the ideal class group.
In a lecture in algebraic number theory, our professor remarked that $$1 \to R^{\times} \to K^{\times} \to \mathcal{I}_K \to \mathcal{C} \ell_K \to 1$$ with $K^{\times} \to \mathcal{I}_K, \ x \mapsto xR$, is exact.
This is not difficult to verify but my question is: So what? What do we learn now that we know that this sequence is exact?
You give the example of an exact sequence defining the class group of a Dedekind domain and you wonder "So what? What do we learn now from it?" At this stage, your question is quite legitimate. Apart from the concision of the language, which allows to conveniently put a certain number of definitions/properties together in a single bunch, what do we gain in comprehension ? Actually this is much more than a simple parlance, this lays the foundations of all the (co)homology theories which have pervaded such numerous and various domains of mathematics as algebraic number theory (and especially class field theory "à la" Artin-Tate), algebraic geometry, algebraic topology, differential geometry, etc. One example being worth a hundred speeches, let us start from a single concrete problem and develop it systematically along the so called cohomological lines.
To stay on familar ground, I pick up your class group short exact sequence for a number field $K$ with ring of integers $O_K$ and group of units $U_K$. It reads : $1\to K^*/U_K\to I_K\to Cl_K\to 1$. Let $L/K$ be a finite galois extension with group $G$. Of course $G$ acts on all the terms of the exact sequence relative to $Cl_L$, and the so called "genus theory" is the study of the natural "extension map" $Cl_K\to {Cl_L}^G$, where $(.)^G$ denotes the subgroup consisting of all the classes of $Cl_L$ fixed by $G$ ("ambiguous ideal classes"). The absolutely automatic first step suggested by the machinery of exact sequences is to take $G$-invariants and get a long exact sequence $1\to (L^*/U_L)^G\to {I_L}^G\to {Cl_L}^G\to ...$ which we must compare with the basic short exact sequence relative to $Cl_K$ in order to "dismantle" the extension map $Cl_K\to {Cl_L}^G$. The process is akin to the study of the behaviour of a differentiable function in the neighbourhood of a given point by using its Taylor expansion. The Taylor expansion is automatic, the true problem actually lies in the interpretation of its coefficients.
In our case, the "coefficients" which come into play are cohomology groups. Recall that (see e.g. Cassels-Fröhlich, ANT, chap.4), starting from an exact sequence of $G$-modules $1\to A\to B\to C\to 1$ and taking $G$-invariants, we get an a priori infinite exact sequence of abelian groups $1\to A^G\to B^G \to C^G\to H^1(G,A)\to H^1(G,B)\to H^1(G,C)\to...\to H^n(G,A)\to H^n(G,B)\to H^n(G,C)...$
Our problem is to express these cohomology groups in terms of arithmetic terms attached to the extension $L/K$. The literature in genus theory being unduly huge, let us cite only a few examples. Denote by $Cap(L/K)$ ("cap" is for "capitulation") and $Cocap(L/K)$ resp. the kernel and cokernel of the extension map $Cl_K\to {Cl_L}^G$. We have exact sequences such as $1\to Cap(L/K)\to H^1(G,U_L)\to {I_L}^G/I_K \to Cocap(L/K)\to H^2(G,U_L)\to Ker(H^2(G, L^*)\to H^2(G, I_L))\to H^1(G, Cl_L)\to H^3(G, U_L)$
or $1\to H^1(G,U_L)\to {I_L}^G/P_K\to {Cl_L}^G\to H^2(G,U_L)\to H^2(G,U_L)\to Ker(H^2(G, L^*)\to H^2(G, I_L))\to H^1(G, Cl_L)\to H^3(G, U_L)$ (where $P_K$ denotes the subgroup of principal ideals of $K$). In particular, if $G$ is cyclic, there is a finite formula giving the quotient of the orders of ${Cl_L}^G$ and $Cl_K$ (Chevalley's formula for ambiguous classes).
In conclusion, let me stress again that there could be no way to "guess" at such results without cohomology.
I was told not to worry about this too much because it really is just a reformulation of the definition of the ideal class group. In particular, this is a short exact sequence $$ 1 \to K^{\times}/R^{\times} \to \mathcal{I}_K \to \mathcal{C}\ell_K \to 1$$ which for me also looks a bit nicer than the sequence with $6$ entries.
