Tensor Notation for Derivative and Covariant Derivative I am learning GR and the notation is killing me here.  So my understanding is, the comma notation is used to indicate a derivative, such as:
\begin{equation}
V^{\alpha}_{\;\;,\gamma}=\partial_{\gamma}V_{\alpha}
\end{equation}
and a semicolon is used to represent a covariant derivative, such as:
\begin{equation}
V^{\alpha}_{\;\;;\gamma}= \partial_{\gamma}V^{\alpha}+\Gamma_{\gamma\mu}^{\alpha}V^{\mu} = V^{\alpha}_{\;\;,\gamma}+\Gamma_{\gamma\mu}^{\alpha}V^{\mu} = \nabla_{\gamma}V^{\alpha}
\end{equation}
However! In problem 7.7 in "The Problem Book of Relativity and Gravitation" they write (for the metric tensor g):
\begin{equation}
g_{\alpha \beta , \gamma} = \nabla_{\gamma}(\mathbf{e}_{\alpha}\cdot \mathbf{e}_{\beta}) = \Gamma^{\mu}_{\alpha\gamma}\mathbf{e}_{\mu}\cdot\mathbf{e}_{\beta}+\Gamma^{\mu}_{\beta \gamma}\mathbf{e}_{\mu}\cdot\mathbf{e}_{\alpha}
\end{equation}
Christoffel symbols?! How? I thought these only popped up when taking the COVARIANT derivative.  Then later on, they write:
\begin{equation}
A^{\alpha}_{\;;\alpha} = A^{\alpha}_{\;,\alpha}+\Gamma^{\alpha}_{\;\beta\alpha}A^{\beta}
\end{equation}
Which makes sense given my definition above, but does not make sense with the notation used in the first example.  Am I missing something? Is it just a typo??
 A: All formulas you shown above are using abstract index notation except the third formula which is fully expressed is a basis.
For a vector field, you can write for example
$$V = V^\mu e_\mu\;,$$
where $V^\mu$ is a scalar while $e_\mu$ is a vector basis. This is a sort of confusion because in abstract index notation we view $V^\mu$ as a vector field.
When we take the covariant derivative, it reads
$$\nabla_\mu V=\nabla_\mu (V^\nu e_\nu) = \nabla_\mu (V^\nu) e_\nu + V^\nu \nabla_\mu ( e_\nu)$$
\begin{eqnarray}
&=& \partial_\mu (V^\nu) e_\nu + V^\nu \Gamma_\mu{}^\lambda{}_\nu e_\lambda\;,\\
&=&\big(  \partial_\mu V^\nu  +\Gamma_\mu{}^\nu{}_\lambda  V^\lambda\big)e_\nu
\end{eqnarray}
If we define 
$$\nabla_\mu V =: (\nabla_\mu V^\nu) e_\nu  $$
we will have the relation in the abstract index notation 
$$\nabla_\mu V^\nu = \partial_\mu V^\nu  +\Gamma_\mu{}^\nu{}_\lambda  V^\lambda\;.$$
(More general, you can start with $\nabla V$ and then define $\nabla V =:(\nabla_\mu V^\nu) e^\mu \otimes e_\nu$  )
Next, the metric $g$, it is (o,2) tensor so it has two slots for inserting 2 vectors if we insert the basis into these slots we will get a component of the metric tensor which is a scalar field
$$g(e_\mu, e_\nu)=g_{\mu\nu}$$
($g= g_{\alpha\beta} e^\alpha \otimes e^\beta,\;g(e_\mu, e_\nu) =g_{\alpha\beta} e^\alpha(e_\mu) \otimes e^\beta(e_\nu) =g_{\alpha\beta}\delta^\alpha_\mu \delta^\beta_\nu= g_{\mu\nu} $)
It is also usually define that $\eta(A,B):= A\cdot B$, $\eta$ is a Minkowskian metric
$A\cdot B$ is a scalar so invariants under coordinate transformations
$$A\cdot B =\eta(A,B) \equiv \eta_{IJ} A^I B^J$$
$$= g(A,B) \equiv g_{\mu\nu} A^\mu B^\nu$$
where $A^I = e^I_\mu A^\mu$ for some scalar $e^I_\mu$ (a vierbein), and you can easily prove that $g_{\mu\nu} = \eta_{IJ} e^I_\mu e^J_\nu$.
So now we have
$$e_\mu \cdot e_\nu =\eta_{IJ}e^I_\mu e^J_\nu= g_{\mu\nu}$$
In this step, we can view $\eta_{IJ},g_{\mu\nu}$ as the scalar fields $e^I_\mu$ as a vector field 
\begin{eqnarray}
\nabla_\gamma g_{\alpha\beta} (= \partial _\gamma g_{\alpha\beta})&=& \nabla_\gamma (e_\alpha\cdot e_\beta) \\
&=&\eta(\nabla_\gamma e_\alpha,e_\beta) +\eta(e_\alpha,\nabla_\gamma e_\beta)
\equiv \eta_{IJ}\nabla_\gamma(e^I_\alpha) e^J_\beta + \eta_{IJ}e^I_\alpha \nabla_\gamma(e^J_\beta) \\
&=&\eta(\Gamma_\gamma{}^\rho{}_\alpha e_\rho,e_\beta) +\eta(e_\alpha,\Gamma_\gamma{}^\sigma{}_\beta  e_\sigma)
\equiv \eta_{IJ}\Gamma_\gamma{}^\rho{}_\alpha e^I_\rho e^J_\beta + \eta_{IJ}e^I_\alpha \Gamma_\gamma{}^\sigma{}_\beta  e^J_\sigma \\
&=&\Gamma_\gamma{}^\rho{}_\alpha \eta( e_\rho,e_\beta) +  \Gamma_\gamma{}^\sigma{}_\beta\eta(e_\alpha, e_\sigma)
\equiv \Gamma_\gamma{}^\rho{}_\alpha \eta_{IJ} e^I_\rho e^J_\beta +  \Gamma_\gamma{}^\sigma{}_\beta \eta_{IJ}e^I_\alpha e^J_\sigma \\
&=& \Gamma_\gamma{}^\rho{}_\alpha  e_\rho \cdot e_\beta +  \Gamma_\gamma{}^\sigma{}_\beta  e_\alpha \cdot e_\sigma \equiv \Gamma_\gamma{}^\rho{}_\alpha g_{\rho \beta} + \Gamma_\gamma{}^\sigma{}_\beta g_{\alpha \sigma}
\end{eqnarray}
Note: Not fully detailed as much as possible but may be helpful for you.
A: This may not be the answer you're looking for because I provide no intuition herein, but from a computational perspective, it's not so hard to see why the derivative of the metric involves Christoffel symbols.  
The affine connection commonly used in general relativity is chosen to be both torsion free and metric compatible.  The second condition means that the covariant derivative of the metric vanishes.
$$
  \nabla_\gamma g_{\alpha\beta} = 0.
$$
These two conditions uniquely specify the connection which is called the Levi-Civita connection.  One can show that the associated covariant derivative of an arbitrary 2-tensor satisfies (see, for example, Carroll's GR, section 3.2):
$$
\nabla_\gamma T_{\alpha\beta} = \partial_\gamma T_{\alpha\beta} - \Gamma_{\gamma\alpha}^\mu T_{\mu\beta} - \Gamma_{\gamma\beta}^\mu T_{\alpha\mu}
$$
Plugging in the metric, noting that the left hand side vanishes, and rearranging gives the desired result.
