I'm not an expert, but I don't find the situation very hard to imagine. It doesn't use anything beyond the fact that "the Lie algebra is the tangent space of the Lie group at the identity". Moreover, for this purpose it is enough to imagine the tangent space as arrows at the identity of the Lie group, pointing in the directions you can move.

Oh, and a side remark that might be helpful to some. Having a map $M \to \mathfrak g^*$ that I usually thought of as a "moment map" is exactly the same as having a map $\mathfrak g \to C^\infty(M,\mathbb R)$ the questioner is talking about, and the latter is easier to visualize.

The simplest case is when $G=\mathbb R$. This is the standard case of the (time-independent) Hamiltonian flow. The Lie-algebra $\mathfrak g$ is one-dimensional, so a linear moment map $f:\mathfrak g \to C^\infty (M,\mathbb R)$ is determined by its value on the unit vector $\vec e \in \mathfrak g$. If $H=f(\vec e)$, this is precisely the Hamiltonian on your manifold. The flow with respect to the moment map is precisely the same as the flow of this Hamiltonian.

The next simplest case is when $G=\mathbb R^n$. The only difference here is that there are several directions in which you could go. Let ${\vec e_1 ,\ldots \vec e_n} \in \mathfrak g$ be the vectors at the identity of $G$ that represent the possible directions. Now, a moment map $f:\mathfrak g \to C^\infty (M,\mathbb R)$ is determined by n Hamiltonians; for each $1\leq i \leq n$ we have $H_i = f(\vec e_i)$.

Now, any path in $G=\mathbb R^n$ will correspond to some flow on the manifold $M$. In particular, if you always go "right" (in the direction of $\vec e_1$), the flow will be precisely that of the Hamiltonian $H_1$; the existence of other directions won't matter. If you only go in the direction of $\vec e_2$, the flow will be that of $H_2$. For any path, you can approximate it by a piecewise-linear path that's always parallel to a coordinate axes; the flow will be that of $H_i$ whenever you go parallel to the axes of $\vec e_i$.

Finally, if G is any Lie group, it's a manifold, so it locally looks like $\mathbb R^n$. Everything I said above about this case still holds, with one exception: the coordinate directions may no longer be "independent". (So, going right, then up, then left, then down, might not bring you exactly to the same point you started) Unfortunately, at this point my expertness runs out, so see the other answers for a detailed explanation. However, algebraically, the condition that you need to add is precisely that the moment map f is a Lie-algebra homomorphism; think of this as an error term you need to add to account for the lack of independence when you change the direction of your piecewise-linear path.

Of course, in practice, in calculating flows you don't need to approximate anything with piecewise-linear paths, as there are algebraic equations will give you the Hamiltonian that corresponds to any direction. In the case of $\mathbb R ^n$, the map f will simply be linear, in general there is likely an error term.

Oh, and finally, most algebraic equations are probably simpler if you think of your moment maps as maps $M\to \mathfrak g^*$.

I'm not an expert, but I don't find the situation very hard to imagine^{(see footnote below)}. It doesn't use anything beyond the fact that "the Lie algebra is the tangent space of the Lie group at the identity". Moreover, for this purpose it is enough to imagine the tangent space as arrows at the identity of the Lie group, pointing in the directions you can move.

Oh, and a side remark that might be helpful to some. Having a map $M \to \mathfrak g^*$ that I usually thought of as a "moment map" is exactly the same as having a map $\mathfrak g \to C^\infty(M,\mathbb R)$ the questioner is talking about, and the latter is easier to visualize.

The simplest case is when $G=\mathbb R$. This is the standard case of the (time-independent) Hamiltonian flow. The Lie-algebra $\mathfrak g$ is one-dimensional, so a linear moment map $f:\mathfrak g \to C^\infty (M,\mathbb R)$ is determined by its value on the unit vector $\vec e \in \mathfrak g$. If $H=f(\vec e)$, this is precisely the Hamiltonian on your manifold. The flow with respect to the moment map is precisely the same as the flow of this Hamiltonian.

The next simplest case is when $G=\mathbb R^n$. The only difference here is that there are several directions in which you could go. Let ${\vec e_1 ,\ldots \vec e_n} \in \mathfrak g$ be the vectors at the identity of $G$ that represent the possible directions. Now, a moment map $f:\mathfrak g \to C^\infty (M,\mathbb R)$ is determined by n Hamiltonians; for each $1\leq i \leq n$ we have $H_i = f(\vec e_i)$.

Now, any path in $G=\mathbb R^n$ will correspond to some flow on the manifold $M$. In particular, if you always go "right" (in the direction of $\vec e_1$), the flow will be precisely that of the Hamiltonian $H_1$; the existence of other directions won't matter. If you only go in the direction of $\vec e_2$, the flow will be that of $H_2$. For any path, you can approximate it by a piecewise-linear path that's always parallel to a coordinate axes; the flow will be that of $H_i$ whenever you go parallel to the axes of $\vec e_i$.

Finally, if G is any Lie group, it's a manifold, so it locally looks like $\mathbb R^n$. Everything I said above about this case still holds, with one exception: the coordinate directions may no longer be "independent". (So, going right, then up, then left, then down, might not bring you exactly to the same point you started) Unfortunately, at this point my expertness runs out, so see the other answers for a detailed explanation. However, algebraically, the condition that you need to add is precisely that the moment map f is a Lie-algebra homomorphism; think of this as an error term you need to add to account for the lack of independence when you change the direction of your piecewise-linear path.

Of course, in practice, in calculating flows you don't need to approximate anything with piecewise-linear paths, as there are algebraic equations will give you the Hamiltonian that corresponds to any direction. In the case of $\mathbb R ^n$, the map f will simply be linear, in general there is likely an error term.

Oh, and finally, most algebraic equations are probably simpler if you think of your moment maps as maps $M \to \mathfrak g^*$.

Footnote: When I wrote that everything is simple, I hadn't thought of the error terms yet (see above). However, I still think that for most conceptual purposes, it's best to think of your Lie group as $\mathbb R^n$ with some error terms added. Somebody correct me if I'm wrong, but when we describe Lie groups using "structure forms", isn't it precisely a way to make this idea precise?