If $\Gamma^k_{ij}(p)=0$, then $\nabla_{E_i}E_j (p)=0?$

That more or less follows from definition. Note that $E_i(q)$ is given by parallel transport along the geodesic $\gamma$ with $\gamma(0) = p$ and $\gamma(1)=q$. Let $\gamma$ be a geodesic such that $\gamma(0) = p$ and $\gamma'(0) = E_i(p)$. Then we have $\gamma'(t) = E_i(\gamma(t))$ and so

$$\nabla_{E_i} E_j (\gamma(t)) = \nabla_{\gamma'} E_j (\gamma(t)) = 0$$

for all $t$, since $E_j$ is by definition the parallel transport along $\gamma$. In particular if you set $t=0$, we have $\nabla_{E_i} E_j (p) = 0$.