For the first problem, you have already detected where the problem lies: the variable $\phi$ is not a function defined on the whole manifold. Indeed, it is a priori a function in a chart on the manifold and a chart usually does not cover by itself the whole manifold.
On the other hand, the particular case of the torus is special because we can more or less canonically 'parametrize' the torus by $\mathbb{R}^2$ (which is its universal cover), for instance via the map $q: \mathbb{R}^2 \to \mathbb{R}^2 / \mathbb{Z}^2 \cong T^2$. As $\phi$ can be chosen to be one of two cartesian coordinates on $\mathbb{R}^2$, its derivative $d\phi$ (on the plane) is left invariant by any translation, in particular the ones by vectors in $\mathbb{Z}^2$. As such, $d\phi$ 'passes to the quotient' i.e. there exists a well-defined closed 1-form $\eta$ on $T^2$ such that $q^{\ast}\eta = d\phi$. This is another motivation to write $\eta = d\phi$, but note that $\phi$ itself would be a multi-valued function on the torus (and hence not a genuine function, so we wouldn't consider it as an antiderivative to $\eta$).
On the sphere, any chart misses at least one point, so again it is not surprising that one can find an antiderivative to a closed 1-form inside this chart. But if you can't extend $\phi$ and $\theta$ to the whole sphere, it is not clear how you can extend their derivatives to globally closed 1-forms in the first place: your problem possibly does not show up. Besides, the fact that the vector field $X = \partial/\partial \theta$ on the sphere can be globally defined (by rotation invariance and also by the null vectors at the poles) is not related to the (im)possibility that $\theta$ (or $d\theta$) is globally well-defined, but only to the fact that $X \lrcorner \omega$ is a closed (and exact) 1-form : an antiderivative is the height function, which is clearly not the angle 'function' $\theta$.
The obstruction to a symplectic vector field $X$ to be Hamiltonian is precisely whether the closed 1-form $X \lrcorner \omega$ is exact. In other words, does the cohomology class $[X \lrcorner \omega] \in H^1_{dR}(M; \mathbb{R})$ vanish? (The nonvanishing of this class is the obstruction to $X$ being Hamiltonian.) This question makes sense on any manifold; the point is that when $H^1_{dR}(M; \mathbb{R})=0$, then the answer is 'yes' whatever the symplectic field $X$. So on the 2-sphere, any symplectic vector field is Hamiltonian, whereas on the torus it depends on the symplectic vector field considered. Put differently, the (non)vanishing of the 1-cohomology group is the obstruction to the equality $Symp(M, \omega) = Ham(M, \omega)$.
I will use the notation $\rho_{t}:=\exp tv$. By assumption, $\rho_{t}$ is the one-parameter group of diffeomorphisms generated by $v$, i.e.
\begin{equation}\tag{1}\label{1}
\begin{cases}
\left.\frac{d}{ds}\right|_{s=t}\rho_{s}(p)=v(\rho_{t}(p))\\
\rho_{0}(p)=p
\end{cases}
\end{equation}
for all $p\in M$ and all $t\in\mathbb{R}$. To see that $g\circ\rho_{t}\circ g^{-1}$ is the one-parameter group of diffeomorphisms generated by $g_{*}v$, we need to show that
\begin{equation}\tag{2}\label{2}
\begin{cases}
\left.\frac{d}{ds}\right|_{s=t}(g\circ\rho_{s}\circ g^{-1})(p)=(g_{*}v)\big((g\circ\rho_{t}\circ g^{-1})(p)\big)\\
(g\circ\rho_{0}\circ g^{-1})(p)=p
\end{cases}
\end{equation}
for all $p\in M$ and $t\in\mathbb{R}$. For the first equality in \eqref{2}, we have
\begin{align}
\left.\frac{d}{ds}\right|_{s=t}(g\circ\rho_{s}\circ g^{-1})(p)&=(dg)_{\rho_{t}(g^{-1}(p))}\left(\left.\frac{d}{ds}\right|_{s=t}(\rho_{s}(g^{-1}(p))\right)\\
&=(dg)_{\rho_{t}(g^{-1}(p))}\left(v(\rho_{t}(g^{-1}(p)))\right)\\
&=\left(g_{*}v\right)\left((g\circ\rho_{t}\circ g^{-1})(p)\right).
\end{align}
Here the first equality is due to the chain rule, and the second equality uses the first equation of \eqref{1} at the point $g^{-1}(p)$.
As for the second equality in \eqref{2}, just use the second equation of \eqref{1} at the point $g^{-1}(p)$.
Best Answer
There are two ways to read your question since you use the phrase "Hamiltonian isotopy", rather than the related notion of "Hamiltonian symplectomorphism"-i.e. a symplectomorphism which Hamiltonian isotopic to the identity.
V1. Is a symplectic isotopy a Hamiltonian isotopy?
No, see Proposition 9.19 of "Introduction to symplectic topology" (Second edition).
V2. Is a symplectomorphism which is isotopic (or even symplectically isotopic) to the identity a Hamiltonian symplectomorphism?
The answer is also no. See see the symplectomorphism of $T^*{S^{1}} $constructed here: Showing that some symplectomorphism isn't Hamiltonian. It is symplectically isotopic to the identity by $F_{t}(x,\xi) = (x,\xi + t)$, but it is not a Hamiltonian symplectomorphism as is proven there.