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 to take into account that the partial integration will also give you contributions from the derivatives of $\sqrt{\det(g_{ij})}$. The formula obtained this way defines a differential operator only on $U$, and a priori the formula might also depend on the coordinates and the trivialization. We now show that the defining equation, i.e. your last displayed equation holds for all $t$ and $s$. This shows the existence of $A^*$ on $U$, denoted in the following as $A_U^*$.
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 from the defining equation that the formal adjoints $A_{U_i}^*$ defined via $U_i$, $i=1,2$ 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.