You prove this as follows: As a first step you show the existence of the formally adjoint differential operator on a small open set $U$. This set $U$ is chosen, such that it carries a chart and such that you can trivialize the bundle $V$ on $U$. Make such a trivialization orthonormal and then write down the equation defining the formally adjoint operator in a chart. The volume form then gives a contribution $\sqrt{\det(g_{ij})}$ as usual. Then standard partial integration in $\mathbb{R}^n$ of the operator $A$ (in coordinates) yields the formally adjoint operator $A^*$ (in coordinates). You have take into account that the partial integration will also give you contributions from the derivatives of  $\sqrt{\det(g_{ij})}$. This shows the existence of $A^*$ on $U$, and its formula might even depend on the choice of $U$, the coordinates, the trivialization, a priori. We then show that the defining equation, i.e. 

Now as second step take two such open sets $U_1$ and $U_2$ with choices of charts and trivializations which we suppress in notation. Then it is easy to see that the formal adjoints $A_{U_i}^*$ defined via $U_i$ agree on $U_1\cap U_2$, more precisely:
On $U_1\cap U_2$ we have  $A_{U_1}^*=A_{U_2}^*$ as differential operators. 

The property in the second step does not only provide uniqueness on small open sets, but it also allows to glue the locally defined operators $A_U^*$ to a globally defined differential operator, which is the requested formally adjoint operator.

Note that this operator is only formally adjoint, it is a different question, whether and how it extends to an adjoint operator, and this extension usually also depends on the type of Sobolev space you use and on possible boundary conditions.