I can't comment on your first question, but for the second one you can just imagine the unit sphere in $\mathbb{R}^3$. Let's parametrize the sphere via
\begin{align*}
X^\mu(\theta, \varphi) = \begin{pmatrix}\sin\theta \cos\varphi \\ \sin\theta \sin\varphi \\ \cos\theta \end{pmatrix} \qquad X: [0, \pi] \times [0, 2\pi] \to \mathbb{R}^3
\end{align*}
The euclidean metric is just the unit matrix $g_{\mu\nu} = \delta_{\mu\nu}$, so the induced metric on the sphere is
\begin{align*}
G_{ab} = \delta_{\mu\nu} \frac{\partial X^\mu}{\partial \sigma^a} \frac{\partial X^\nu}{\sigma^b} &= \sum_{\mu=1}^3 \frac{\partial X^\mu}{\partial \sigma^a} \frac{\partial X^\mu}{\partial \sigma^b} \\
&= \begin{pmatrix}\cos^2\theta \cos^2\varphi + \cos^2\theta \sin^2\varphi + \sin^2\theta & \cos\theta\cos\varphi\sin\theta (-\sin\varphi) + \cos\theta\sin\varphi \sin\theta\cos\varphi \\ \cos\theta\cos\varphi\sin\theta (-\sin\varphi) + \cos\theta\sin\varphi \sin\theta\cos\varphi & \sin^2\theta\sin^2\varphi + \sin^2\theta\cos^2\varphi \end{pmatrix} \\
&= \begin{pmatrix} 1 & 0 \\ 0 & \sin^2\theta \end{pmatrix}
\end{align*}
Now the determinant is simply $\det G = \sin^2\theta$ and so the volume of the sphere is
\begin{align*}
V(\mathbb{S}^2) = \int\mathrm{d}^2\sigma\ \sqrt{|\det G |} = \int_0^\pi\mathrm{d}\theta \int_0^{2\pi}\mathrm{d}\varphi\ \sin\theta = 2\cdot 2\pi = 4\pi
\end{align*}
as expected.
Hopefully someone else can flesh the following out a bit more, but here's a rough sketch on why you might expect the square root of the determinant of the metric: the volume-form is completely antisymmetric $\mathrm{d}\sigma^1 \dots \mathrm{d}\sigma^n = \frac{1}{n!} \varepsilon_{\rho_1 \dots \rho_n} \mathrm{d}\sigma^{\rho_1} \dots \mathrm{d}\sigma^{\rho_n}$. Under coordinate transformations the Levi-Civita symbol will turn the derivatives from the chain-rule into a Jacobian factor
\begin{align*}
\frac{1}{n!} \varepsilon_{\rho_1 \dots \rho_n} \mathrm{d}\sigma^{\rho_1} \dots \mathrm{d}\sigma^{\rho_n} = \frac{1}{n!}\det\left(\frac{\partial \sigma}{\partial \tilde{\sigma}}\right)\varepsilon_{\kappa_1 \dots \kappa_n}\mathrm{d}\tilde{\sigma}^{\kappa_1}\dots \mathrm{d}\tilde{\sigma}^{\kappa_n} = J\ \mathrm{d}\tilde{\sigma}^1 \dots \mathrm{d}\tilde{\sigma}^n
\end{align*}
The metric transforms as
\begin{align*}
G_{a' b'} = \frac{\partial \sigma^a}{\partial\tilde\sigma^{a'}} \frac{\partial \sigma^b}{\partial \tilde{\sigma}^{b'}} G_{ab} \qquad\Rightarrow\qquad \det G' = \det\left(\frac{\partial \sigma}{\partial \tilde\sigma}\right)^{2}\ \det G = J^2\ \det G
\end{align*}
By combining the two transformations we find an invariant integrand
\begin{align*}
\int\mathrm{d}^n\sigma\ \left(\det G\right)^{1/2} = \int J\,\mathrm{d}^n\tilde\sigma\ \left(J^{-2} \det G' \right)^{1/2} = \int\mathrm{d}^n\tilde\sigma\ \left(\det G'\right)^{1/2}
\end{align*}
