This is a non-trivial fact. It is possible to create surfaces of revolution of the form
$$(\phi(v) \cos(u), \phi(v) \sin(u), \psi(v)),\quad \phi \ne 0$$
If $v$ is arc length we have $(\phi')^2 + (\psi')^2 = 1$. Having constant Gaussian curvature, K, then implies $\phi''+K\phi = 0$ and $\psi = \int \sqrt{1-(\psi')^2}\ dv$.
With $K=1$ and the curve intersection the X-Y plane perpendicularly we have
$$\phi(v) = C \cos v,\quad \psi(v) = \int_0^v \sqrt{1-C^2 \sin^2 v}\ dv.$$
So $\psi(v)$ is an Incomplete elliptic integral of the second kind. (Reference Do Carmo)
Now these surfaces have constant positive Gaussian curvature, if $C=1$, it gives a sphere, if $C\ne 1$, you have surface which have two singular points on the rotation axis. Examples of such surfaces can be seen at Wolfram demonstrations.
One of the comments above points to a looseness in Wikipedia's statement. The sphere is the only compact, simply-connected, Riemannian surface of constant positive curvature. The above examples, either fails to be compact if we exclude the singular points, or if they are left it fails to be smooth, hence not Riemannian.
Wikipedia does give a hint to a proof of the fact, it references the Liebmann's theorem (1900), and a brief comment that
A standard proof uses Hilbert's lemma that non-umbilical points of extreme principal curvature have non-positive Gaussian curvature.
This may have come from Hilbert and Cohn Vossen (p228). They first show that surface of constant positive Gaussian curvature, without boundary or singularities, must be a closed surface. Apart from the sphere there are no surfaces where both principal curvatures are constant. So we just need to consider cases like the above where the two principle curvatures vary, but their products are constant. As the surface is closed there must be a point where one of the principle curvature obtains its maximum value. But it can be proved analytically that such points can't exist, except on the boundary, on surfaces with constant positive Gaussian curvature.
do Carmo, Manfredo Perdigão, Differential geometry of curves and surfaces. Mineola, NY: Dover Publications (ISBN 978-0-486-80699-0/pbk). xvi, 510 p. (2016). ZBL1352.53002.
Best Answer
The quick way to get the correct answer is dimensional analysis: to get an area from a dimensionless angle defect you need to multiply by something with units length$^2$, so Wikipedia's formula $A = (\pi - \sum\theta)R^2$ is correct.
Sommerville is a very old text and uses $k$ for the negative inverse curvature - see e.g. page 75 where it states "The formulae of hyperbolic trigonometry become of those of euclidean plane geometry as $k \to \infty$."
The proof is going to depend on how you are constructing things - from the Riemannian perspective you just calculate how curvatures and areas behave under multiplication of the metric by a constant: covariant derivatives are unchanged, so Ricci curvature is unchanged, so scalar curvature scales as the inverse metric. Area scales as the metric. Thus area is inversely proportional to curvature under scaling.