So I wrote up this small derivation drawing insights from the answer by Deane Yang and Steve Huntsman,

I have checked that the following text compiles without any errors otherwise but here on MO some of the lines aren't compiling!

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 $S^n$ !

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 $S^n$ !

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 metric $g_{\mu \nu}$ 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 metric 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,

$$\nabla _{\mathbb{R}^{n+1}}^2 \phi = \frac{n}{r}\partial _r \phi + \partial _r ^2 \phi + \frac{\nabla _{S^n} ^2 \phi}{r^2}$$

So I wrote up this small derivation drawing insights from the answer by Deane Yang and Steve Huntsman,

I have checked that the following text compiles without any errors otherwise but here on MO some of the lines aren't compiling!

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 $S^n$ !