Why are projective objects important?

I think the main reason for the interest in projective modules is the extraordinary bijective correspondence discovered by Serre between finitely generated projective modules $P$ on a completely arbitrary ring $A$ and locally free sheaves of finite rank (= vector bundles) $\mathcal F$ on the corresponding affine scheme $X=Spec(A)$. The correspondence simply associates to a vector bundle $\mathcal F$ the $A$-module of its global sections: $P=\Gamma(X, \mathcal F)$.

This gives a dictionary between algebra and geometry in which free modules correspond to trivial bundles, etc. Swan and Forster have perfected this dictionary by bringing topology and complex analysis onto the scene. A particularly interesting application of that dictionary is that one could prove that the projective module over the ring of functions of the 2-sphere corresponding to the tangent bundle of that sphere is not trivial by appealing to the well known topological fact that the tangent bundle is topologically non trivial. No proof had been available before.

Maybe the most spectacular succcess story of the dictionary is the simultaneous but independent proof by Quillen and Suslin in 1977 that every vector bundle on affine space $\mathbb A^n_k=Spec(k[T_1,\ldots,T_n])$ over a field $k$ is trivial by showing that finitely generated projective bundles over $k[T_1,\ldots,T_n]$ are trivial . This answered a question Serre had asked on page 243 of the article in which he introduced his correspondence: Faisceaux Algébriques Cohérents, always lovingly called FAC . This article is one of the most important ones in twentieth century mathematics. (English translation here, found thanks to MathOverflow)

Edit At Zev's request, let me try to explain why vector bundles give rise to projective modules . I'll do it in the more familiar context of topology, the idea being the same as in algebraic geometry. Given a vector bundle $F$ on a compact topological space $X$, it is easy to see that there is a finite number of continuous sections $s_1, \ldots,s_N\in \Gamma(X,F)$ which generate the fiber of $F$ at each $x\in X$, even though $\Gamma(X,F)$ itself is (almost) always infinite dimensional.This yields a surjective morphism of vector bundles $\pi : \theta^N\to F$ from the rank-$N$ trivial bundle on $X$ to $F$.
The kernel of $\pi$ is then a vector bundle $K$ on $X$ and the exact sequence of bundles $0\to K \to \theta^N \to F\to 0$ splits (they all do on compact spaces !) and we obtain a direct sum decomposition $\theta ^N=K\oplus F$. Taking global sections we obtain $$C(X)^N= \Gamma(X,K) \oplus \Gamma(X,F)$$ from which projectivity ( and finite generation) of $ \Gamma(X,F) $ over $C(X)$ follows

If I've understood Pete Clark's answer to another question correctly, here is a basic reason: in $R\text{-Mod}$ ($R$ a commutative ring) the dualizable objects are precisely the finitely-generated projective modules. For a good discussion of the importance of dualizable objects see, for example, Ponto and Shulman's Traces in symmetric monoidal categories.

Projective objects $P$ are important because they are the ones for which $\hom(P, -)$ is an exact functor - it is usually only half exact. Similarly, injective objects are the ones for which $\hom(-,I)$ is exact. This means that the derived functors (Ext-functors) of $\hom(P,-)$ and $\hom(-,I)$ vanish.

If you are given a ring, it is sometimes quite non-obvious how to construct its modules.

One natural place to look for modules is inside the ring itself and inside the free modules, which are immediately constructible from the ring itself. Now, inside free modules there can be all sort of things... but it is quite apparent that the simplest of those modules we are going to find inside free modules are precisely the summands thereof.

(1) In terms of representation theory, an $R$-module $M$ determines a representation of the ring $R$:

$R \rightarrow \text{End}(M)$

A projective $R$-module $P$ corresponds to a sub representation of a standard representation of $R$ on a free $R$-module $M = \prod_{i \in I} R$ that is also a direct summand: $M = P\oplus Q$. Thus, the map

$R \rightarrow \text{End}(M) \stackrel{\text{Res}_{P}}{\rightarrow} \text{Hom}(P, M) = \text{Hom}(P, P\oplus Q) = \text{End}(P) \oplus \text{Hom}(P, Q)$,

factors through

$R \rightarrow \text{End}(P)$.

Thus, once we know which free module $P$ is a direct summand of, we immediately know how $R$ acts on it.

(2) Linear functionals $P \rightarrow R$ on a projective $R$-module $P$ separate points, i.e. for all $p, q, \in P$, there exists a linear functional $f$ such that $f(p) \neq f(q)$. Because $f$ is linear, we can assume that $q = 0$, and so this is equivalent to the following: for every $p \in P$, there exists $f \in \text{Hom}_R(P, R)$, such that $f(p) \neq 0$. From this we see that this condition precisely says that the canonical map

$P \stackrel{\varphi}{\rightarrow} (P^\ast)^\ast$

$p \mapsto \varphi_p : (f \mapsto f(p))$

is injective, i.e. $\varphi_p = 0$, if, and only if, $f(p) = 0$ for all $f \in \text{Hom}_R(P, R)$, if, and only if, $p = 0$. Now, to show that $\varphi$ is injective, note that there is an $R$-module $Q$ such that $M = P\oplus Q$ is free. Suppose that $p\neq 0$. Let $f \in \text{Hom}_R(M, R)$, such that $f(p) \neq 0$, and restrict $f$ to $P$ (i.e. just take $f$ to be the projection onto a non-zero component of $p$). Then $\phi_p(f) \neq 0$. Thus, $\phi$ is injective.

More generally, an $R$-module $M$ is called torsionless if the canonical map $M \rightarrow (M^\ast)^\ast$, is injective. Thus projective modules are torsionless.