To give an answer for the sake of it being here.
The Ricci tensor is defined as usual, as the trace of the Riemann curvature tensor.
Some authors use Ricci curvature to denote the function on the unit tangent bundle:
$$ UTM \ni v \mapsto \mathrm{Ric}(v,v)\in \mathbb{R} $$
The values of this function is sometimes called the Ricci curvatures.
Constant Ricci curvature at a point $p$ means that for all $v\in UT_pM$ we have $\mathrm{Ric}(v,v) = c$ for some constant $c$. In three dimensions because the Weyl curvature vanishes identically the Riemann curvature tensor is uniquely determined by the Ricci tensor, and the Ricci tensor being symmetric is uniquely determined by the Ricci curvature function. Hence you can in principle write the Riemann curvature tensor in terms of the Ricci curvature (by polarisation) and show that the sectional curvatures are constant.
Or you can use that in $\mathbb{R}^3$ the space of two-dimensional subspaces is itself three dimensional, so one can actually solve a linear system of equations to write sectional curvatures in terms of Ricci curvatures.
You can observe that by Gauss Equation and Codazzi-Mainardi equation you can write the Riemann tensor and the Ricci tensor in function of the second fundamental form of your hypersurface that is in $\mathbb{R}^{n+1}$, so the Riemann and Ricci tensor of the ambient are zero.
The Weyl tensor is defined in function of the induced metric, the Riemann tensor and the Ricci tensor of the hypersurface so it is true, it is possible write the Weyl tensor in function of the second fundamental form.
We know that Gauss equation of an Hypersurface $M\subseteq N$ is
$R^N(X,Y,Z,W)=R^M(X,Y,Z,W)+h(X,W)h(Y,Z)-h(X,Z)h(Y,W)$
where $h$ is the second fundamental form of $M$.
In local coordinates we have that
$R^N_{ijkl}=R^M_{ijkl}+h_{il}h_{jk}-h_{ik}h_{jl}$
In our case $R^N=0$ so
$R^M_{ijkl}= h_{ik}h_{jl}-h_{il}h_{jk}$
If we trace the equation we can write also Ricci Tensor in function of the second fundamental form:
$R^M_{jl}=Hh_{jl}-g^{ik} h_{il}h_{jk}$
where $H$ is the mean curvature of $M$.
If we trace twice Gauss equation we get a relationship between the scalar curvature $R$ of $M$ and the second fundamental form:
$R=H^2-|h|^2$
We consider the (0,4)-Weyl tensor of $M$
$W=R^M-\frac{1}{n-2}(Ric-\frac{R}{n}g)\wedge g-\frac{R}{2n(n-1)}g\wedge g$
where $\wedge$ is the Kulkarni-Nomizou product.
In local coordinate we have that
$W_{ijkl}=R^M_{ijkl}+\frac{1}{n-2}(R^M_{il}g_{jk}-R^M_{ik}g_{il}+R^M_{jk}g_{il}-R^M_{jl}g_{ik})+$
$+\frac{R}{(n-1)(n-2)}(g_{ik}g_{jl}-g_{il}g_{jk})$
So you have all elements to write Weyl tensor in function of the second fundamental form of $M$ (and in function of the metric $g$).
You can observe also that the (0,2)-Tensor $h$ is symmetric so there exists a local orthonormal frame $\{Z_1,\dots, Z_n\}$ such that the matrix $A$ associated with $h$ with respect that frame is $diag(\lambda_1,\dots, \lambda_n)$ and th matrix $G$ associated with the metric $g$ with respect that frame will be the identity $I$. In this case you can write all our formula in an easier way than before.
If it is not clear I can explain more precisely.
Best Answer
There is an error in the computation of the Ricci curvature, but without more details it's not possible to say where.
From OP's comment under the question it appears that $\gamma$ is an arbitrary metric on $S^2$. The coordinates are local, so this question is more or less just asking about the behavior of the cone metric $$\hat \gamma := dr^2 + r^2 \pi^* \gamma$$ over a Riemannian surface $(\Sigma, \gamma)$ on $\hat\Sigma := \Bbb R_+ \times \Sigma$. Here, $\pi : \hat\Sigma = \Bbb R_+ \times \Sigma \to \Sigma$ is just the canonical projection onto the second factor. Computing directly$^*$ gives that the respective Ricci curvatures $\operatorname{Ric}, \widehat{\operatorname{Ric}}$ of $\gamma, \hat\gamma$ are related by $$\widehat{\operatorname{Ric}} = \pi^*(\operatorname{Ric} - \gamma) ,$$ and then forming traces gives that the respective scalar curvatures $R, \hat R$ are related by $$\hat R = r^{-2} \pi^* (R - 2) .$$ These two identities immediately give that $$\widehat{\operatorname{Ric}} = 0 \Leftrightarrow \operatorname{Ric} = \gamma \Leftrightarrow \hat R = 0 \Leftrightarrow R = 2 ,$$ which in particular recovers Yu Ding's observation from the comments.
$^*$One can compute these identities efficiently using isothermal coordinates: There are some local coordinates $(x_1, x_2)$ on $\Sigma$ for which $\gamma = e^{2 \Upsilon} (dx_1^2 + dx_2^2)$ for some smooth function $\Upsilon(x_1, x_2)$. Applying the Koszul Formula yields relations between $\hat\nabla$ and $\nabla$, e.g., $$\hat \nabla_{\partial_{x_i}} \partial_{x_j} = \nabla_{\partial_{x_i}} \partial_{x_j} - r \gamma(\partial_{x_i}, \partial_{x_j}) \partial_r ,$$ among simpler relations, and then compute the curvatures directly from these. (In this display equation, we've implicitly used the decompositions $T_{(r, p)} \hat\Sigma = T_r \Bbb R_+ \oplus T_p \Sigma$.)