Can anyone give me a reference which explain the derivation of the partial differential operator expression for the laplacian on the euclidean n-dimensional space and on $S^n$ ?
One generally writes the laplacian on the n-dim euclidean space as a sum of a operator on the radial coordinate and $\frac{1}{r^2}$ times the laplacian on $S^n$.
And very often the laplacian on $S^n$ is written through a recursion relation.
I am looking for a reference which shows me the derivations of these.
The Laplacian can be defined on any Riemannian manifold as div grad. Here grad f for f a smooth function is the vector field dual to the 1-form df via the bilinear form of the metric. Div of a vector field X corresponds to taking the covariant derivative $\nabla X$, which is a (1,1) tensor, and taking the trace of that. In local coordinates one can give a formula using the symbols for the metric, which should yield what you are looking for.
Another way to define div is to take the Lie derivative of the volume form: that is, $L_X V = (div X) V$. The volume form depends on an orientation, which can be locally chosen. This way is actually probably easier for computing in local coordinates since you don't need to worry about a covariant derivative or Christoffel symbols.
For a reference, see e.g. Taylor's Partial Differential Equations, Vol. 1. In Folland's Introduction to Partial Differential Equations, there isn't much about Riemannian manifolds, but Folland does talk about how the Laplacian changes with respect to new coordinates.
The Laplacian originates from physics. In particular, it arises as the linear differential operator in the Euler-Lagrange equation for the functional $f \mapsto E[f] = \int |\nabla f|^2$. You can derive formulas for the Laplacian on either Euclidean space or the unit sphere by differentiating this functional with respect to $f$ and determining the condition for a critical point.
You can figure out the relationship between the Euclidean and spherical Laplacians by observing that in polar co-ordinates, $|\nabla f|^2 = |\partial_rf|^2 + r^2|\partial_\theta f|^2$, where $\nabla$ is the Euclidean gradient and $\partial_\theta$ is the spherical gradient.
The recurrence relation for the spherical Laplacian arises from the observation in polar co-ordinates the $(n-1)$-dimensional spherical gradient can be written as $|\partial_\theta f|^2 = |\partial_\phi f|^2 + (\sin\phi)^2|\partial'_\theta f|^2$, where $\phi \in [0,\pi)$ is the co-ordinate giving the angle between a point and $e_n$, $\theta \in S^{n-2}$, and $\partial_\theta f$ is the $(n-2)$-dimensional spherical gradient.
These formulas, at least for dimensions 2 and 3, can be found in most textbooks on electromagnetic theory or mathematical physics. The trick is to use the same derivation given in these books but recast them in a more abstract arbitrary dimension form.
If you want to work things out using Riemannian geometry, I recommend using stereographic co-ordinates on the unit sphere.
In $\mathbb{R}^n$ consider $f = f(r)$. Writing $\partial_j \equiv \partial/\partial x_j$ and $\partial_r \equiv \partial/\partial_r$, etc., we have that $\partial_j r = x_j/r$, so
$\partial_j f = (\partial_r f)(x_j r^{-1})$
and
$\partial_{jj} f = (\partial_{rr} f)(x_j^2 r^{-2}) + (\partial_r f)(r^{-1} - x_j^2 r^{-3})$.
Summing over $j$ and comparing with the Cartesian expression for $\Delta$ gives the decomposition into radial and spherical operators. To be explicit you should consider $f = f(r, \omega)$, where $\omega \in S^{n-1}$.
For a more general case, see the end of Chapter 2 of The Laplacian on a Riemannian manifold: an introduction to analysis on manifolds by Rosenberg. Unfortunately some of the relevant section (the Laplacian in exponential coordinates) is blocked both from Amazon and Google:
I recommend Terras, Harmonic analysis on symmetric spaces.
So I wrote up this small derivation drawing insights from the answer by Deane Yang and Steve Huntsman,
With respect to the Riemann-Christoffel connection on a riemannian manifold the laplacian on that manifold will have the form $$\nabla ^2 \phi = \frac{1}{\sqrt{g}} \partial _{\mu} \left [ \sqrt{g} g^{\mu \nu} \partial _{\nu} \phi \right ]$$
where $g$ is the the determinant of the metric on the manifold and $\phi$ is some smooth scalar function on the manifold.
On can write the line element on $S^n \subset \mathbb{R}^{n+1}$ as,
$d\Omega _n ^2 = d\theta _1 ^2 + sin^2 \theta_1 d\theta _2 ^2 + sin^2 \theta _1 sin^2 \theta_2 d\theta _3 ^3 +...+sin^2\theta _1 sin^2 \theta_2...sin^2 \theta_{n-2} sin^2 \theta_{n-1} d\theta _n ^2$
Then the line element on $\mathbb{R}^{n+1}$ in polar coordinates can be written as,
$$ds^2 = dr^2 + r^2 d\Omega _n ^2$$
and $g_{\mathbb{R}^{n+1}} = r^{2n}g_{_{S^n}}$ where $g_{_{S^n}} = (sin^2 \theta _1)^{n-1}(sin^2 \theta_2)^{n-2}...(sin^2 \theta _{n-2})^2(sin^2 \theta_{n-1})^1$
Therefore since the metric is diagonal $\nabla _{\mathbb{R}^{n+1}} ^2 \phi = \frac{1}{r^n \sqrt{g_{_{S^n}} }} \partial _{\mu} \left [r^n \sqrt{g_{_{S^n}} } g^{\mu \mu}_{\mathbb{R}^{n+1}} \partial _{\mu} \phi \right ]$
$=\frac{1}{r^n \sqrt{g_{_{S^n}} }} \partial _{r} \left [r^n \sqrt{g_{_{S^n}} } \partial _{r} \phi \right ] + \frac{1}{r^n \sqrt{g_{_{S^n}} }} \partial _{\theta _i} \left [r^n \sqrt{g_{_{S^n}} } g^{\theta _i \theta _i}_{\mathbb{R}^{n+1}} \partial _{\theta _i} \phi \right ]$
$=\frac{1}{r^n}\partial _r (r^n \partial_r \phi) + \frac{1}{\sqrt{g_{_{S^n}} }} \partial_{\theta _i} \left[ \sqrt{g_{_{S^n}} } \frac{g^{\theta _i \theta _i}_{S^n}} {r^2} \partial _{\theta _i} \phi \right]$
$=\frac{1}{r^n}\partial _r (r^n \partial_r \phi) + \frac{\nabla _{S^n}^2 \phi}{r^2}$
Therefore after doing the differentiation we have the final result,
$$\nabla _{\mathbb{R}^{n+1}}^2 \phi = \frac{n}{r}\partial _r \phi + \partial _r ^2 \phi + \frac{\nabla _{S^n} ^2 \phi}{r^2}$$
And I don't see an neat way of writing the Laplacian on $S^n$ !
