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.
Best Answer
In OP's last $\approx$ sign, the index on the covariant derivatives $\nabla \varphi$ should be lowered by use of the inverse metric. This yields 2 minus contributions, so that the corrected eq. (2) agrees with OP's eq. (1).
See also this related Phys.SE post.