I'll give hints instead of a complete solution, in case this ends up being categorized as homework.
I'll write $|g|$ instead of $-g$, because the former works for either sign convention (mostly-plus or mostly-minus). The quantity
$$
V^\mu := \frac{\partial {\cal L}}{\partial(\partial_\mu\phi)}
\tag{1}
$$
is a vector (I mean, the components of a vector), so the question is how to derive
$$
\frac{1}{\sqrt{|g|}}\partial_\mu\left(\sqrt{|g|} V^\mu\right)
=\nabla_\mu V^\mu
\tag{2}
$$
for a vector $V$. Use the definition of $\nabla_\mu$ to see that (2) can also be written
$$
\frac{1}{\sqrt{|g|}} V^\mu\partial_\mu\sqrt{|g|}
= \Gamma^{\alpha}_{\alpha\mu}V^\mu.
\tag{3}
$$
The left-hand side of (3) involves the quantity
$$
\frac{1}{\sqrt{|g|}}\partial_\mu\sqrt{|g|}
=
\frac{1}{2}\partial_\mu\log|g|.
\tag{4}
$$
We can think of $g$ as a square matrix, and $|g|$ is its determinant. The matrix is invertible, so $|g|$ is the product of its eigenvalues $\lambda_n$. Rewrite the right-hand side of (4) in terms of $\lambda_n$, and recall that if $A,B$ are diagonalizable square matrices, then the trace of their product is equal to the sum of the products of their eigenvalues. Use this write the right-hand side of (4) in terms of the matrices $g^{-1}$ and $\partial_\mu g$. Written in terms of indices, this becomes
$$
\frac{1}{\sqrt{|g|}}\partial_\mu\sqrt{|g|}
=\frac{1}{2}g^{\alpha\beta}\partial_\mu g_{\alpha\beta}.
\tag{5}
$$
We can use this to rewrite the left-hand side of (3). Now write the $\Gamma^{\alpha}_{\alpha\mu}$ on the right-hand side of (3) in terms of the metric and its derivatives to see that the identity (3) is indeed true.
Reference: Chapter 3 in Ward (1984), General Relativity, University of Chicago Press. Equation (3.4.10) in that book is equation (3) above.
