I am trying to do some calculations with involving the Ricci tensor, but I am off by a factor 2. Here's my problem: The manifold is $S^2$ with the standard metric $g=e^{2\phi}\delta_{ij}$, with $\phi=\ln2-\ln(1+x^2+y^2)$. Using formulas from this wikipedia page I have $$\Gamma^1_{11}=-\Gamma^1_{22}=\Gamma^2_{12}=\phi_{,1}$$ and $$\Gamma^1_{12}=-\Gamma^2_{11}=\Gamma^2_{22}=\phi_{,2}.$$ If I understand correctly, the Ricci tensor is $$R_{ij}=g_{ij}$$ since the radius of my sphere is 1. I want to calculate $\nabla_m\nabla_iY^m$. Using the Ricci tensor, I would expect
$$
\nabla_m\nabla_iY^m=\nabla_i\nabla_mY^m+R_{im}Y^m =(d(div(Y)))_i+(Y^\flat)_i.
$$
I want to test this formula with a simple vector field $Y^m=x\delta^m_2-y\delta^m_1$ (with, as usual $x=x^1$ and $y=x^2$). I have $$ \nabla_iY^j=\frac{x^2+y^2}{1+x^2+y^2}(\delta^2_i\delta^j_1-\delta^1_i\delta^j_2)=A(x,y) (\delta^2_i\delta^j_1-\delta^1_i\delta^j_2). $$ It follows that $$div(Y)=\nabla_iY^i=0.$$ Also $$ \nabla_m\nabla_iY^j=A_{,m}(\delta^2_i\delta^j_1-\delta^1_i\delta^j_2)+A(-\Gamma^2_{mi}\delta^j_1+\Gamma^j_{1m}\delta^2_i+\Gamma^1_{mi}\delta^j_2-\Gamma^j_{m2}\delta^1_i) $$ Contracting $$ \nabla_m\nabla_iY^i=A(-\Gamma^2_{1m}+\Gamma^2_{1m}+\Gamma^1_{m2}-\Gamma^1_{m2}), $$ which is to be expected since $div(y)=0$, and $$ \nabla_m\nabla_iY^m=A_{,1}\delta^2_i-A_{,2}\delta^1_i+A(-\Gamma^2_{i1}+\Gamma^m_{1m}\delta^2_i+\Gamma^1_{2i}-\Gamma^m_{2m}\delta^1_i). $$ Using the calculated values for the $\Gamma$'s, the above expression simplifies $$ \nabla_m\nabla_iY^m=-A_{,2}\delta^1_i+A_{,1}\delta^2_i=\frac{1}{2}g_{im}Y^m=\frac{1}{2}(Y^\flat)_i. $$ which is half of what I would expect based on formula based on the Ricci tensor. What am I doing wrong?
Too long for another comment.
Your metric can be written as \begin{align} g_{11}&=g_{22}=\frac{4}{(1+x^2+y^2)^2}\,, &g_{12}&=g_{21}=0\,. \end{align} I verified with a symbolic calculator that in this case the Ricci tensor $R_{ij}$ is indeed equal to $g_{ij}$ and the scalar curvature is two like we get on the unit sphere from the standard metric that I mentioned in a comment.
So the question arises: What were your coordinates and where does this exercise come from?
Another remark on your calculation of Christoffel symbols. Do you know the difference between $\Gamma_{ijk}$ and $\Gamma^i_{ij}\,?$ In calculations you start with the former, and in your case \begin{align} \Gamma_{111} &= -\frac{8x}{1+x^2+y^2}\\ \end{align} which happens to be equal $\frac \partial{\partial x}\phi$ in your special case.
You expect an identity for the covariant directional derivative $\nabla_m\nabla_iY^m\,.$
Proof. By the Ricci identity in your link, $$ \nabla_k\nabla_iY^m-\nabla_i\nabla_kY^m=-R_{ki\ell}{}^mY^\ell $$ where $R_{ki\ell}{}^m$ is the Riemann tensor which is antisymmetric in $ki\,$. Since the Ricci tensor is the contracted Riemann tensor $R_{k\ell}=R_{km\ell}{}^m$ we get $$ \nabla_m\nabla_iY^m-\nabla_i\nabla_mY^m=R_{i\ell}Y^\ell\,. $$