Short answer:
When you do the Jordan-Wigner transformation, you essentially insert a linear combination of tensor products of Pauli matrices for each fermionic creation and annihilation operator. As there is a one-to-one correspondence (you're basically just "rewriting" what $a_p^\dagger$, $a_r$ stand for on the quantum computer), it is irrelevant whether you do this before or after trotterization.
Long answer:
In second quantization, in general, each $h_i$ is a linear combination of creation and annihilation operators
$$
h_i=\sum_{pq\dots rs} c_{i,pq\dots rs} a_p^\dagger a_q^\dagger \dots a_r a_s
$$
where $p,q,\dots,r,s$ each go from 1 to $M$, where $M$ is the number of fermionic modes.
Now, performing the Jordan-Wigner mapping, each operator $a_p^\dagger$, $a_r$ is replaced by a linear combination of tensor products of Pauli spin matrices. We thus get a "quantum computer representation" of the operator $h_i$.
It does not matter at all when we perform the Jordan-Wigner transformation. As $a_p^\dagger$, $a_r$ will be replaced by a linear combination of tensor products of Pauli spin matrices, it is just a question of notation whether you do that before or after trotterization.
On a side note: Operators which are local acting on fermionic systems (acting on a constant number of fermionic modes, e.g. $a_p^\dagger a_r$ acts on two fermionic modes) are not necessarily local when represented on a quantum computer (e.g. $a_p^\dagger a_r$ acts on $O(M)$ qubits, see for example here). As $h_i$ should be local operators, performing the Jordan-Wigner transformation might turn it into a highly non-local operator. There are other valid transformations between second quantized operators and qubit operators that get $O(M)$ down to $O(log(M))$ (most notably, the Bravyi-Kitaev transformation). From that point of view, it does indeed matter when to perform a transformation (and which to perform) - but that has more to do with computational efficiency.
