Gradient-like vector fields

No. Since your question is about a neighborhood of a critical point, we can work over $\mathbb{R}^n$ instead of the compact manifold $M$.

Consider $\mathbb{R}^2$ with the following two coordinate charts in a neighborhood of 0. First we have the standard $x,y$ coordinates. Next we have the coordinates

$$ z = x \cos r^2 + y \sin r^2 \qquad w = y \cos r^2 - x \sin r^2 $$

where $r^2 = x^2 + y^2$. We easily verify that $z^2 + w^2 = x^2 + y^2 = r^2$. So that both $(x,y)$ and $(z,w)$ are Morse charts for $f = r^2$.

Let the vector field $X$ be $- x\partial_x - y\partial_y$ in the $(x,y)$ coordinates, and $X'$ be $- z\partial_z - w\partial_w$ in the $(z,w)$ coordinates. You can compute the change of variables explicitly and see that $X \neq X'$ except at the origin.

(It may be easier to see in standard polar coordinates, where $X = r\partial_r$ and $X' = r\partial_r + 2r^2\partial_\theta$. With this you also see that by adding a cut-off at finite $r$ for the perturbation, we can also directly extend this example to any two dimensional manifold. Higher dimensional analogues are also immediate.)