I believe the following way (Kostant's, 1970) to be the best way to think about the Hamiltonian condition.
First, "why" is there a central extension $H^0(M; {\mathbb R}) \to C^\infty (M) \to symp(M)$ of Lie algebras? Of what is $C^\infty (M)$ supposed to be the Lie group? For $symp(M)$, the Lie algebra of vector fields annihilating the symplectic form $\omega$, it's clear it should the Lie algebra of the group $Symp(M)$ of symplectomorphisms.
Assume now that $[\omega]$ is integral. Then it is $c_1$ of some "prequantization" line bundle $\mathcal L$, and $\omega$ is the curvature of some Hermitian connection $\alpha$ on that line bundle. Let $Aut(M,{\mathcal L},\alpha)$ denote the group of Hermitian bundle automorphisms (moving the base around) of $\mathcal L$ preserving $\alpha$. This group obviously maps to $Diff(M)$, forgetting the action on the fibers, but because it preserves $\alpha$ on $\mathcal L$ it preserves $\omega$ on $M$, so the image lies inside $Symp(M)$. The kernel consists of bundle automorphisms that only act fiberwise, and for them to preserve the flat connection they must, on each component, rotate all the fibers by the same element of $U(1)$.
The Hamiltonian condition, then, is about whether one can lift the action of $G$ on $M$ to an action on the line bundle over $M$. It's very easy, given such a lift, to write down a moment map. (Basically, now that you're dealing with a $1$-form $\alpha$ instead of a $2$-form $\omega$, you can pair vector fields from $\mathfrak g$ with it.)
One example I find instructive is ${\mathbb R}^{2n}$ acting on itself by translation, with the space given the usual symplectic structure. That's acting as symplectomorphisms, and the space is simply connected, so there's no $H^1$ obstruction (as when $T^1$ acts on $T^2$). But one can't lift the action to preserve the (non-flat) connection on the (trivial) line bundle; it only lifts to an action of the Heisenberg group.
Another subtle example is $SO(3)$ acting on $S^2$ with the area $1$ symplectic structure. On the Lie algebra level, yes, the action is Hamiltonian. But actually $SO(3)$ doesn't act on the line bundle; only its double cover $SU(2)$ does.
Finally, think about the case that $G$ acts algebraically on $X \subseteq {\mathbb P}V$. I like to say that $X$ is "equivariantly projective" if $G$ acts on ${\mathbb P}V$ preserving $X$, and this is pretty nearly an algebro-geometric replacement for the Hamiltonian condition. (Non-example: $X$ is a nodal cubic curve, whose smooth locus is ${\mathbb C}^\times$, acted on by ${\mathbb C}^\times$.)