Let $M = \mathbb{R} - \{0\}$ and $X$ be the vector field $d/dx$ on $M$. Find the maximal integral curve of $X$ starting at $x=1$.
So we are looking for $c : (a,b) \to \mathbb{R} - \{0\}$ such that $c'(t) = X_{c(t)}$ for all $t \in (a,b)$.
Since $c$ is a map to $\mathbb{R} - \{0\}$ we can write it as $c(t) = x(t)$ and so $c'(t) = \dot{x}(t)$. We now have to get $$d/dx = X_{c(t)} = c'(t) = \dot{x}(t).$$
How can we solve such an differential equation with the initial condition $c(0) = 1$? I'm confused with the different derivatives some being considered with the manifold and some in $\mathbb{R} - \{0\}$.
The notation may be creating some confusion here. The vector field $X$ on $M = \Bbb R - \{0\}$ is $\frac{d}{dx}$, which is the restriction of the vector field with the same name on $\Bbb R$. Unwinding definitions shows that, the velocity vector field along a smooth curve $c: (a, b) \to M$ is $$t \mapsto \dot c(t) \left.\frac{d}{dx}\right\vert_{c(t)}.$$ So, by definition $c$ is an integral curve of $X$ if $$\dot c(t) \left.\frac{d}{dx}\right\vert_{c(t)} = X_{c(t)} = \left.\frac{d}{dx}\right\vert_{c(t)} ,$$ or just as well (by comparing coefficients with respect to the frame $\left(\frac{d}{dx}\right)$) if $$\dot c(t) = 1$$ (for all $t$), and the initial condition is, as you write, $c(0) = 1$. Can you solve this initial value problem?