If there is a chart $(U,\phi)=(U,x^1,\dots,x^n)$ on $M$ with center $p$ and a tangent vector $\langle a^1,\dots,a^n\rangle$ with respect to the bases $\left\{\frac{\partial}{\partial x^i}\right\}$, then a curve $c_1$ starting at $p$ such that $c_1'(0)=X_p$ can be generated by defining $\alpha(t)=(a^1t,\dots,a^nt)$ and letting $c_1=\phi^{-1}\circ\alpha$.
If there is another chart $(V,\psi)=(V,y^1,\dots,y^n)$ then for the same tangent vector $X_p$ with a different coordinate form with respect to the new bases $\left\{\frac{\partial}{\partial y^i}\right\}$, there will be a new curve $c_2$ generated by the same method.
My question is whether $c_1=c_2$.
No. For a really simple example, consider rectangular and polar coordinates on $\mathbb{R}^2$. Starting at $(1,0)$ (in rectangular coordinates), the curve given in rectangular coordinates given by a vertical velocity vector is a vertical line. But in polar coordinates, it is a circle about the origin (since a vertical velocity vector points in the $\theta$ direction).
What's going on here is that on a manifold, it doesn't make sense to talk about tangent vectors at different points as being "the same". On $\mathbb{R}^n$ we can identify the tangent space at every point with $\mathbb{R}^n$ and thus given just a single vector at a point we can turn it into a "constant" vector field defined everywhere and consider an integral curve for that vector field. But on a manifold, you can't take a tangent vector at a single point and turn it into a vector field in a canonical way. You can do this using a single coordinate chart, but different coordinate charts will give different results.