The short answer is that $z$ is a dependent variable and $x$ and $y$ are independent variables. Therefore, the Jacobian must be between $x$ and $y$ and $\phi$ and $\theta$, without $z$.
This is similar to what the OP suggested, just set up in a different way. Consider the function:
$$
z=f(x,y)=\sqrt{4-x^2-y^2}.
$$
We can see, directly, that $z\geq 1$ means that $4-x^2-y^2\geq 1$ or that $3\geq x^2+y^2$. This describes the disk centered at the origin of radius $\sqrt{3}$. Let $D$ be the disk of radius $\sqrt{3}$.
The surface area formula is
$$
\iint_D\sqrt{f_x(x,y)^2+f_y(x,y)^2+1}dA.
$$
In this case,
\begin{align}
f_x&=\frac{-x}{\sqrt{4-x^2-y^2}}&f_y&=\frac{-y}{\sqrt{4-x^2-y^2}}.
\end{align}
Therefore, the surface area integral becomes
$$
\iint_D\sqrt{\frac{x^2}{4-x^2-y^2}+\frac{y^2}{4-x^2-y^2}+1}dA
=
\iint_D\sqrt{\frac{4}{4-x^2-y^2}}dA=\iint_D\frac{2}{\sqrt{4-x^2-y^2}}dA.
$$
Here's the $2$ that you found in your computation.
Now, let's switch to polar, the disk $D$ is defined by $0\leq\theta\leq 2\pi$ and $0\leq r\leq \sqrt{3}$, so, by switching to polar, we have
$$
\int_0^{2\pi}\int_0^{\sqrt{3}}\frac{2r}{\sqrt{4-r^2}}drd\theta=\int_0^{\sqrt{3}}\frac{4\pi r}{\sqrt{4-r^2}}dr
$$
Now, use the $u$-substitution $u=4-r^2$ to get your final answer.
If you were to compute the determinant that you need to convert $x$ and $y$ to $\theta$ and $r$ (remember that $z$ is a dependent variable in your setup), you should get the same formula.
With spherical coordinates, as you set up, you'll get a somewhat more complicated formula by substituting in your formulae for $x$ and $y$ in terms of $\phi$ and $\theta$ and taking the determinant of the change of variables from $x$ and $y$ to $\phi$ and $\theta$. The integral that you should get is
$$
\int_0^{\pi/3}\int_0^{2\pi}\frac{8|\cos(\theta)\sin(\theta)|}{\sqrt{4-4\sin^2(\theta)}}d\phi d\theta=
\int_0^{\pi/3}\frac{16\pi|\cos(\theta)\sin(\theta)|}{\sqrt{4-4\sin^2(\theta)}}d\phi
$$
where the absolute value of the determinant of the Jacobian is $4|\cos(\theta)\sin(\theta)|$.
