On the method of characteristics By the method of characteristics, it is possible to prove the existence of local solution for the first-order linear non homogeneous equation:

$\mathcal{L}_X f+g=0$

where $X$ is a non-singular vector field on a manifold $M$, $g$ is a given function on $M$, and $f$ is the unknown one.
Question: What could I say on the existence of solutions for the similar equation:

$\mathcal{L}_{X_t} f+g_t=0$,

with the only difference that now the non-singular vector field $X_t$ and the function $g_t$ on $M$ are time-dependent, while I need a time-independent solution?
Could it be useful to know that, in the problem I am tackling, I need a solution around a point $m_0\in M$ where $X_t(m_0)=X_0(m_0)\neq 0$ and $g_t(m_0)=0$ for all $t$?
 A: Rather then thinking of $t$ as time, you should think of it as a free parameter. So your question is asking, whether given a family of differential equations, there can be one simultaneous solution to all of them. Seen this way, the answer is clearly no in general. Even with your additional conditions. 
As an example, consider for arbitrary fixed $\epsilon$, consider the time dependent family of vector fields
$$ X(t) = \partial_x + (2\epsilon^2 x - 4x^3) t \partial_y $$
on $\mathbb{R}^2$. Observe that near $m_0 = (0,0)$ your condition on $X$ is satisfied. Let the source function
$$ g(t) = g(0) \begin{cases} > 0 & y < 0 \\ = 0 & y \geq 0 \end{cases} $$
Observe that for every $t$ there exists an integral curve of $X(t)$ connecting the origin $(0,0)$ to the point $(\epsilon,0)$. For $t < 0$ this curve, between the two point, lies entirely in the lower half plane. For $t \geq 0$, the corresponding segment lies in the closed upper half plane.
So we get a contradiction. Let $f_0 = f(0,0)$ of the purported solution. For any $t\geq 0$, solving the equation requires $f(\epsilon,0) = f_0$. But for any $t < 0$, solving the equation requires the strict inequality $f(\epsilon,) < f_0$. 

Remember that the method of characteristics reduces to essentially solving ordinary differential equations along integral curves. Non-uniqueness as described above will always be a possible problem if in your family there exists two different integral curves (corresponding to different parameters $t_1$ and $t_2$) connecting the same two points. In fact, the situation is slightly worse even then that:
To ensure the existence of a solution, a necessary condition is a holonomy type constraint. That is, you need that for any two points $p,q\in M$, for any piecewise integral curve $\gamma$ connecting $p$ to $q$ (that is, there exist $s_0 < s_1 < s_2 < \cdots < s_n$ such that $\gamma(s_0) = 0$, $\gamma(s_n) = q$, and parameters $t_1,\ldots, t_n$ such that $\gamma|_{(s_{i-1},s_i)}$ is an integral curve for $X_{t_i}$), the integral $\int_p^q g = \sum_{i = 1}^n \int_{s_{i-1}}^{s_i} g_{t_i}(s) \mathrm{d}s$ is independent of the choice of constituent integral curve segments. 
And even in this case you cannot guarantee that the solution exists for the initial value problem. In the usual case with the method of characteristics, on any non-characteristic hypersurface $\Sigma$ we can prescribe arbitrary initial data and solve the IVP. In the case you described, the family of parametrised vector fields can lead to there being a non-local constraint being necessary for data prescribed on a hypersurface $\Sigma$ that is simultaneously non-characteristic to all of the $X_t$. 
