Commenting on Peter Michor's answer, I want to say that we can however use the exponential map.

Is it true that the image of the exponential map is locally dense near the origin? If so, let us take open neighbourhoods $U\subset V$ of the identity such that $\overline{\exp(Vect_c(M))}\supset \overline{V}$. Given any $g\in V$, we can find $h_0\in U$ arbitrarily close to $id$ so that $gh_0^{-1}=f$ is the time $1$ map of a flow $\{f_t\}_{t\in[0,1]}$. Let $k=k(f)$ be the minimum integer such that $f_{1/k}$ belongs to $U$. Then $g$ is the product of $k+1$ diffeomorphisms in $U$: $$g=f_{1/k}\circ\cdots\circ f_{1/k}\circ h_0.$$

The problem here is that $\overline{U}$ is not compact so it doesn't seem clear that we can have a uniform bound on the choice of $k$ (though we still have some freedom in the choice of $h_0$).

Edit Katie Mann is true that the image of the exponential map can happen to be not dense about the origin. There is one case for which I know that my question has a negative answer and this is a result by Nancy Kopell in Commuting diffeomorphisms, Global Analysis, Proc. Symmpos. Pure Math. vol XIV (1968), 165-184, see also Yoccoz, Petits diviseurs en dimension 1 Astérisque 231, SMF (1995)):

Kopell's Theorem states that the $C^1$ centralizer is trivial for an open dense set of $C^1$ circle diffeomorphisms. In particular there exists an open dense set of diffeomorphisms of the circle which are not the time 1 map of a flow.

Commenting on Peter Michor's answer, I want to say that we can however use the exponential map.

Is it true that the image of the exponential map is locally dense near the origin? If so, let us take open neighbourhoods $U\subset V$ of the identity such that $\overline{\exp(Vect_c(M))}\supset \overline{V}$. Given any $g\in V$, we can find $h_0\in U$ arbitrarily close to $id$ so that $gh_0^{-1}=f$ is the time $1$ map of a flow $\{f_t\}_{t\in[0,1]}$. Let $k=k(f)$ be the minimum integer such that $f_{1/k}$ belongs to $U$. Then $g$ is the product of $k+1$ diffeomorphisms in $U$: $$g=f_{1/k}\circ\cdots\circ f_{1/k}\circ h_0.$$

The problem here is that $\overline{U}$ is not compact so it doesn't seem clear that we can have a uniform bound on the choice of $k$ (though we still have some freedom in the choice of $h_0$).

Edit Katie Mann is true that the image of the exponential map can happen to be not dense about the origin. There is one case for which I know that my question has a negative answer and this is a result by Nancy Kopell in Commuting diffeomorphisms, Global Analysis, Proc. Symmpos. Pure Math. vol XIV (1968), 165-184 (AMS page - subscription required), see also Yoccoz, Petits diviseurs en dimension 1 Astérisque 231, SMF (1995) (Numdam):

Kopell's Theorem states that the $C^1$ centralizer is trivial for an open dense set of $C^1$ circle diffeomorphisms. In particular there exists an open dense set of diffeomorphisms of the circle which are not the time 1 map of a flow.

Commenting on Peter Michor's answer, I want to say that we can however use the exponential map.

Is it true that the image of the exponential map is locally dense near the origin? If so, let us take open neighbourhoods $U\subset V$ of the identity such that $\overline{\exp(Vect_c(M))}\supset \overline{V}$. Given any $g\in V$, we can find $h_0\in U$ arbitrarily close to $id$ so that $gh_0^{-1}=f$ is the time $1$ map of a flow $\{f_t\}_{t\in[0,1]}$. Let $k=k(f)$ be the minimum integer such that $f_{1/k}$ belongs to $U$. Then $g$ is the product of $k+1$ diffeomorphisms in $U$: $$g=f_{1/k}\circ\cdots\circ f_{1/k}\circ h_0.$$

The problem here is that $\overline{U}$ is not compact so it doesn't seem clear that we can have a uniform bound on the choice of $k$ (though we still have some freedom in the choice of $h_0$).

Commenting on Peter Michor's answer, I want to say that we can however use the exponential map.

Is it true that the image of the exponential map is locally dense near the origin? If so, let us take open neighbourhoods $U\subset V$ of the identity such that $\overline{\exp(Vect_c(M))}\supset \overline{V}$. Given any $g\in V$, we can find $h_0\in U$ arbitrarily close to $id$ so that $gh_0^{-1}=f$ is the time $1$ map of a flow $\{f_t\}_{t\in[0,1]}$. Let $k=k(f)$ be the minimum integer such that $f_{1/k}$ belongs to $U$. Then $g$ is the product of $k+1$ diffeomorphisms in $U$: $$g=f_{1/k}\circ\cdots\circ f_{1/k}\circ h_0.$$

The problem here is that $\overline{U}$ is not compact so it doesn't seem clear that we can have a uniform bound on the choice of $k$ (though we still have some freedom in the choice of $h_0$).

Edit Katie Mann is true that the image of the exponential map can happen to be not dense about the origin. There is one case for which I know that my question has a negative answer and this is a result by Nancy Kopell in Commuting diffeomorphisms, Global Analysis, Proc. Symmpos. Pure Math. vol XIV (1968), 165-184, see also Yoccoz, Petits diviseurs en dimension 1 Astérisque 231, SMF (1995)):

Kopell's Theorem states that the $C^1$ centralizer is trivial for an open dense set of $C^1$ circle diffeomorphisms. In particular there exists an open dense set of diffeomorphisms of the circle which are not the time 1 map of a flow.