You forgot to check $F$ is well-defined since it is apparently dependent on the path you choose. First of all cover $\Omega$ by open balls. We shall show there exists a primitive on each ball.
Let $B(z_0,R)$ be a ball. For $z\in B(z_0,R)$, choose the radial path from $z_0$ to $z$, call that $\gamma_z$.
Next define $$F(z)=\int_{\gamma_z}f(\xi)d\xi$$
Then we observe that for $h$ small, $$\frac{1}{h}[F(z+h)-F(z)]=\frac{1}{h}\int_{L(z,z+h)}f(\xi)d\xi$$by Gousrat Theorem where $L(z,z+h)$ is the straight line joining $z$ to $z+h$
Then we get $$\frac{1}{h}[F(z+h)-F(z)]=\frac{1}{h}\int_0^1f(z+\theta h)hd\theta=\int_0^1f(z+\theta h)d\theta \rightarrow f(z)$$as $h\rightarrow 0$ by your favourite convergence theorem.
Thus we have showed the existence of an anti-derivative on an open ball.
This in particular shows that integral of a holomorphic function on a closed curve in any open ball is 0.
Now let $H$ be a fixed end point homotopy between two paths $\gamma_0,\gamma_1$ in a region $\Omega$
Say $H: I^2\rightarrow \Omega$
Choose a partition of $I^2$ into a grid $\{G_{ij}\}$ so that any small tile $G_{ij}$ falls into an open ball in $\Omega$ via $H$ using continuity and compactness. Join the corners of the tiles in $\Omega$ by straight lines and for the sake of simplicity call them $G_{ij}$ as well.
Then one can write $$\int_{\gamma_0}f(\xi)d\xi -\int_{\gamma_1}f(\xi)d\xi =\sum_{i,j}\int_{\partial G_{ij}} f(\xi) d\xi $$
Each term in the last sum is $0$ as it is the integral of a holomorphic function on an open ball by our choice of the partition.
This shows integral of a holomorphic function on 2 fixed end-point homotopic curves is the same. In particular this shows integral of a holomorphic function on a closed curve in any simply-connected domain is 0.
Then you can proceed as you were doing.
If you are interested in a purely algebraic topology approach, here's one way to proceed.
We have solved the primitive problem locally an open cover say balls $\mathcal B=\{ B_i\}$. Any $2$ such solutions on a ball differ by a constant. Say we fix a local solution $\{f_i \}_i$ on the local cover $B_i$
Then on the intersection $B_i\cap B_j$ we get a complex number $c_{ij}$ such $f_i-f_j=c_{ij}$. Thus we get a co-cycle in the Cech cohomology group $\hat {H^1}(\Omega ;\mathcal B)$
The cover we chose was a Leray cover as intersections of convex sets are convex and hence all contractible. So the obstruction is an element of $\hat{H^1}(\Omega; \mathbb C)\cong {H_{dR}^1}(\Omega; \mathbb C)$
For simply-connected smooth manifolds, the $1$st de-Rham cohomology group is $0$ and hence we get the obstruction we got isn't an obstruction at all.
No. Consider this modified version of the Warsaw circle, which has two instances of the Topologist's sine curve on it instead of just one. This shape can be given in polar coordinates as $r(\theta)=\cos\left(e^{\tan(\theta)}\right)+2$. Taking the open set $O$ as the interior region cut out by this curve, $O$ is bounded, simply connected, and regular, but its boundary is the closed image of this modified warsaw circle, which is not path connected. There is no way to path connect across the singularities at $\theta=\pm\frac{\pi}{2}$, so the boundary is path disconnected.
Best Answer
You know that $\gamma \in \mathbb{C}$ can be seen as a curve in $\mathbb{R}^2$. Using the Schonflies statement of the Jordan theorem, there is a homeomorphism $f: \mathbb{R}^2 \to \mathbb{R}^2$ that sends $\gamma$ to the unit circle, and your $W$ gets sent to the open unit disk. It will send $\Omega$ to some open set around the unit disk. Let us construct a set $V$ so that $f(\gamma \cup W) \subseteq V \subseteq f(\Omega)$. Let $\epsilon_\theta$ be the radial distance along the angle $\theta$ between $\gamma$ and the first time the ray intersects $\partial f(\Omega)$. We can define $\epsilon = \inf \epsilon_\theta$.
Notice that this infimum cannot be $0$. This would occur if there were a sequence of points on $\partial f(\Omega)$ that approached a point on $f(\gamma)$, but then this point of $f(\gamma)$ would be in $\partial(f(\Omega))$ since $\partial(f(\Omega))$ is closed. This can't happen because $\Omega$ is open so $int( f(\Omega)) = f(\Omega)$, so if $f(\gamma)$ had a point in the $\partial(f(\Omega))$, then this would contradict the fact that $\gamma \subset \Omega$.
Now let $V$ be the open unit disk of radius $1 + \epsilon$, and let $\Omega' = f^{-1}(V)$. Since this is an open disk, it is simply connected.