Let me describe a way to construct the desired measure directly giving uniformity, which is, in my opinion, much more elegant than using Riemannian geometry or the Hausdorff measure. It also gives a simple characterisation directly proving that it is equivalent to the other constructions. If you are just interested in the question why the Hausdorff measure is finite and non-trivial, you can jump to the last paragraph.

The sphere $S^n$ is homeomorphic to the quotient space $SO(n+1)/SO(n)$, where $SO(n)$ denotes the special orthogonal group. We can regard $S^n$ as a homogeneous space on which $SO(n+1)$ acts by left-multiplication. $SO(n+1)$ is a compact group thus there exists a unique normalised Haar measure on $SO(n+1)$. By compactness of the subgroup $SO(n)$ (actually by agreement of the modular function of the Haar measure of $SO(n+1)$ and $SO(n)$) we get that there exists a unique normalised Radon-measure on $S^n$ invariant under the action of $SO(n+1)$. Since all the groups are comact, the measure is particularly easy to define: Let $p\colon SO(n+1)\to SO(n+1)/SO(n)=S^n$ be the canonical surjection. For every continuous function $f\in C(S^n)$ we can define the integral as

$$
\int_{S^n} f = \int_{SO(n+1)} f\circ p
$$

inducing the measure we are interested in. We did not regard $S^n$ as a metric space yet, but if we do by regarding it as the $L^2$ unit sphere embedded into $R^{n+1}$, then the action of $SO(n+1)$ can be described as rotation of the space which is isometrically and transitively on $S^n$ (and $SO(n)$ is just the isotropy group of a point in $S^n$, using this observation the isomorphy of $SO(n+1)/SO(n)$ and $S^n$ regarded as homogeneous spaces follows). We get the desired result.

For a relation to the Hausdorff measure consider the homeomorphism projecting the $n$-dimensional unit disk onto a hemisphere of the $n$-sphere. This map is Lipschitz with a Lipschitz constant $\pi$. The Hausdorff measure of the (flat) unit disk is well-known and finite, thus the Hausdorff measure of the hemisphere is finite (since Lipschitz functions can only increase Hausdorff measure by a constant factor) and we get finiteness of the Hausdorff measure of the sphere. The inverse of the homeomorphism is a contraction, thus the Hausdorff measure of the hemishere is non-zero. We get that the Hausdorff measure is a non-trivial $SO(n+1)$-invariant Radon measure, thus it is (up to a constant) the same as the measure constructed above.