= Solution
Use the nonnegative sign convention $\Delta=-\operatorname{div}\operatorname{grad}$ for the <Laplace-Beltrami operator>; the ambient <Laplacian> has the same sign convention. Changing both signs changes the signs of the <eigenvalues> below. Let $e_1,\ldots,e_n$ be an <orthonormal basis> of $T_pM$, and let $\gamma_i$ be the unit-speed <geodesic> with $\gamma_i(0)=p$ and $\dot\gamma_i(0)=e_i$. The <geodesic Hessian formula> gives
$$
\boxed{\Delta f(p)=-\sum_{i=1}^n\left.\frac{d^2}{dt^2}f(\gamma_i(t))\right|_{t=0}.}
$$
Indeed, the second derivative along a <geodesic> is $\operatorname{Hess}f(e_i,e_i)$, because its covariant acceleration is zero. Taking the <trace> of the <Hessian matrix> in this <orthonormal basis> gives the displayed <Laplace-Beltrami operator>.
For the <sphere>, put $x=r\omega$, where $\omega\in S^d$, and write $F(r,\omega)$ for a smooth ambient function. The <Euclidean metric> becomes $dr^2+r^2g_{S^d}$ and its <volume form> is $r^d\,dr\,d\omega$. The coordinate formula for the <Laplace-Beltrami operator> therefore gives the <polar-coordinate Laplacian identity>
$$
\widetilde\Delta F=-\partial_r^2F-\frac d r\partial_rF+\frac1{r^2}\Delta_{S^d}F.
$$
For $f(\omega)=F(1,\omega)$, restriction to $r=1$ yields
$$
\boxed{\Delta_{S^d}f=\left(\widetilde\Delta F+\partial_r^2F+d\partial_rF\right)\big|_{r=1}.}
$$
This relation also has a direct <geodesic> proof. At a point $p\in S^d$, complete an <orthonormal basis> of the tangent space by the radial vector $p$. Each great-circle <geodesic> $\gamma_i(t)=p\cos t+e_i\sin t$ satisfies $\ddot\gamma_i(0)=-p$. The chain rule gives $(F\circ\gamma_i)''(0)=D^2F(e_i,e_i)-\partial_rF$. Summing and comparing with the ambient <trace> gives precisely the radial correction above.
Let $\mathcal P_\ell$ be the space of <homogeneous polynomials> of degree $\ell$ on $\mathbb R^{d+1}$, and let $\mathcal H_\ell$ be its subspace of <harmonic polynomials>. For $H\in\mathcal H_\ell$, homogeneity gives $H(r\omega)=r^\ell H(\omega)$. Substituting in the <polar-coordinate Laplacian identity> proves that its restriction $h$ is a <spherical harmonic> with
$$
\Delta_{S^d}h=\ell(\ell+d-1)h.
$$
Restriction is injective on $\mathcal H_\ell$: if $h=0$, homogeneity makes $H$ zero away from the origin, hence everywhere.
To count and exhaust these <eigenfunctions>, use the <harmonic decomposition of homogeneous polynomials>
$$
\mathcal P_\ell=\mathcal H_\ell\oplus r^2\mathcal P_{\ell-2},\qquad \mathcal P_j=0\text{ for }j<0.
$$
Here is an algebraic proof rather than an assumption about the <spectrum>. On <polynomials> put the <Fischer inner product> $\langle x^\alpha,x^\beta\rangle_F=\alpha!\,\delta_{\alpha\beta}$. Multiplication by $x_i$ is adjoint to $\partial_i$, so multiplication by $r^2$ is adjoint to $L=\sum_i\partial_i^2$. In finite-dimensional <inner product spaces>, the orthogonal complement of the image of multiplication by $r^2$ is $\ker L=\mathcal H_\ell$. This proves the decomposition. Multiplication by $r^2$ is injective, so
$$
\dim\mathcal H_\ell=\binom{\ell+d}{d}-\binom{\ell+d-2}{d},
$$
where the second term is zero for $\ell<2$.
Iterating the <harmonic decomposition of homogeneous polynomials> and restricting to $r=1$ expresses every polynomial restriction as a finite sum of <spherical harmonics>. Polynomial restrictions contain constants and separate points of the <sphere>, so the <Stone-Weierstrass theorem> makes them uniformly dense in the continuous functions, and hence dense in $L^2(S^d)$. The <Laplace-Beltrami operator> is <self-adjoint>, and <eigenfunctions> with different <eigenvalues> are <orthogonal>. If a smooth <eigenfunction> had an <eigenvalue> not in the list, it would be orthogonal to a dense subspace and would vanish. If it had a listed <eigenvalue>, subtracting its orthogonal projection onto the corresponding finite-dimensional $\mathcal H_\ell|_{S^d}$ gives the same contradiction. Thus, for $d\geq1$, the <spectrum of the Laplacian on a sphere> is
$$
\boxed{\lambda_\ell=\ell(\ell+d-1),\quad
m_\ell=\binom{\ell+d}{d}-\binom{\ell+d-2}{d},\quad \ell=0,1,2,\ldots.}
$$
In particular, the zero <eigenvalue> has multiplicity one; on $S^1$ each positive <eigenvalue> $\ell^2$ has multiplicity two; on $S^2$ the multiplicity is $2\ell+1$.
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