Here is a partial answer. Observe that whenever you have a (right) action $\theta$ of $G$ on $M$, your Lie algebra morphism $\hat{\theta} \colon \mathfrak{g} \to \mathfrak{X}(M)$ (it is usually called the infinitesimal generator of $\mathbf{\theta}$) sends every $X$ to a complete vector field (because this vector field comes from some global flow on $M$).

Necessary condition: A Lie algebra morphism $\mathfrak{g} \to \mathfrak{X}(M)$ must send every element of $\mathfrak{g}$ to a complete vector field in order to be an infinitesimal generator of some (right) action of $G$ on $M$.

This condition is also sufficient if $G$ is simply-connected:

Let $G$ be a simply-connected Lie group with a Lie algebra $\mathfrak{g}$, and let $\phi \colon \mathfrak{g} \to \mathfrak{X}(M)$ be a Lie algebra morphism sending every element of $\mathfrak{g}$ to a complete vector field. The there is a unique smooth (right) action of $G$ on $M$ whose infinitesimal generator is $\phi$.

See Theorem 20.16 in Introduction to Smooth Manifolds by John Lee.

Here is a partial answer. Observe that whenever you have a (right) action $\theta$ of $G$ on $M$, your Lie algebra morphism $\hat{\theta} \colon \mathfrak{g} \to \mathfrak{X}(M)$ (it is usually called the infinitesimal generator of $\mathbf{\theta}$) sends every $X$ to a complete vector field (you may check that the flow of $\hat{\theta}(X)$ is given by: $(t, p) \mapsto p \cdot exp(tX)$; this flow is definitely global, which means that the vector field is complete).

Necessary condition: A Lie algebra morphism $\mathfrak{g} \to \mathfrak{X}(M)$ must send every element of $\mathfrak{g}$ to a complete vector field in order to be an infinitesimal generator of some (right) action of $G$ on $M$.

This condition is also sufficient if $G$ is simply-connected:

Let $G$ be a simply-connected Lie group with a Lie algebra $\mathfrak{g}$, and let $\varphi \colon \mathfrak{g} \to \mathfrak{X}(M)$ be a Lie algebra morphism sending every element of $\mathfrak{g}$ to a complete vector field. The there is a unique smooth (right) action of $G$ on $M$ whose infinitesimal generator is $\varphi$.

See Theorem 20.16 in Introduction to Smooth Manifolds by John Lee.

Here is a partial answer. Observe that whenever you have a (right) action $\theta$ of $G$ on $M$, your Lie algebra morphism $\hat{\theta} \colon \mathfrak{g} \to \mathfrak{X}(M)$ (it is usually called the \bfseries{infinitesimal generator of $\theta$}) sends every $X$ to a \textit{complete} vector field (because this vector field comes from some \textit{global} flow on $M$).

\textbf{Necessary condition:} A Lie algebra morphism $\mathfrak{g} \to \mathfrak{X}(M)$ must send every element of $\mathfrak{g}$ to a complete vector field in order to be an infinitesimal generator of some (right) action of $G$ on $M$.

This condition is also sufficient if $G$ is simply-connected:

Let $G$ be a simply-connected Lie group with a Lie algebra $\mathfrak{g}$, and let $\phi \colon \mathfrak{g} \to \mathfrak{X}(M)$ be a Lie algebra morphism sending every element of $\mathfrak{g}$ to a complete vector field. The there is a unique smooth (right) action of $G$ on $M$ whose infinitesimal generator is $\phi$.

See Theorem 20.16 in Introduction to Smooth Manifolds by John Lee.

Here is a partial answer. Observe that whenever you have a (right) action $\theta$ of $G$ on $M$, your Lie algebra morphism $\hat{\theta} \colon \mathfrak{g} \to \mathfrak{X}(M)$ (it is usually called the infinitesimal generator of $\mathbf{\theta}$) sends every $X$ to a complete vector field (because this vector field comes from some global flow on $M$).

Necessary condition: A Lie algebra morphism $\mathfrak{g} \to \mathfrak{X}(M)$ must send every element of $\mathfrak{g}$ to a complete vector field in order to be an infinitesimal generator of some (right) action of $G$ on $M$.

This condition is also sufficient if $G$ is simply-connected:

Let $G$ be a simply-connected Lie group with a Lie algebra $\mathfrak{g}$, and let $\phi \colon \mathfrak{g} \to \mathfrak{X}(M)$ be a Lie algebra morphism sending every element of $\mathfrak{g}$ to a complete vector field. The there is a unique smooth (right) action of $G$ on $M$ whose infinitesimal generator is $\phi$.

See Theorem 20.16 in Introduction to Smooth Manifolds by John Lee.