Add the elements of two linked lists in a third list

I see you use a doubly linked list so you should have no problem traversing the numbers from least significant to most significant.

Proposal

I would advise you to pass the list to a number and then let the system be the one in charge of the carries and overflows; the function to pass your Lista to number would be:

int lista_a_numero(const Lista &lista)
{
    int resultado{}, arrastre{1};

    for (const nodo *actual = ultimo(lista); actual; actual = actual->ant)
    {
        resultado += (actual->numero * arrastre);
        arrastre *= 10;
    }

    return resultado;
}

With this function you could transform the list into a number. If you notice, I'm using a function to point to the last node in a list, which looks like this:

const nodo *ultimo(const nodo *actual)
{
    if (actual->sig)
        return ultimo(actual->sig);

    return actual;
}

Now you only need a function that does the reverse operation to get this:

int valor1 = lista_a_numero(lista1);
int valor2 = lista_a_numero(lista2);
Lista listaSuma = numero_a_lista(valor1 + valor2);

The function to transform from number to list would be something like this:

Lista numero_a_lista(int numero)
{
    nodo *resultado = new nodo{numero % 10};
    numero /= 10;

    while (numero)
    {
        resultado = new nodo{numero % 10, resultado};
        resultado->sig->ant = resultado;
        numero /= 10;
    }

    return resultado;
}

With this proposal the code would look like this:

int main()
{
    Lista lista1 = numero_a_lista(689);
    Lista lista2 = numero_a_lista(574);

    int valor1 = lista_a_numero(lista1);
    int valor2 = lista_a_numero(lista2);
    Lista listaSuma = numero_a_lista(valor1 + valor2);

    std::cout << valor1 << " + " << valor2 << " = ";

    for (nodo *actual = listaSuma; actual; actual = actual->sig)
        std::cout << actual->numero;

    return 0;
}

[Whose departure is] :

689 + 574 = 1263

Tips

In C ++ it is not necessary to prepend struct to refer to a structure ; so your node would look like this:

struct nodo
{
 int numero;
 nodo* sig;
 nodo* ant;
};

From C ++ 11 you can use the uniform initialization to give value to the members of your structure in the same site that they declare , so you save yourself creating a constructor and you ensure you get controlled values:

struct nodo
{
 int numero = 0;
 nodo* sig = nullptr;
 nodo* ant = nullptr;
};

To create aliases of types, the typedef has a confusing syntax, from C ++ 11 you can simplify the alias with the instruction using :

using Lista = nodo *;

Note that it does not require the struct before nodo .

As for this alias, it is a terrible idea , nefarious, horrible, inadequate, improper, illogical and many other negative qualifiers. A nodo is not a Lista , conceptually they are different things ... it's as if you were saying that a spark plug is a car; create an object Lista do not say that a nodo is a Lista .

Use const always do not modify objects , that can help the compiler to make optimizations and make others who read the code know your intentionality:

// mostrar una lista no requiere modificarla
void mostrar(const Lista lista);
// no se requiere modificar lista1 ni lista2 para sumar
void sumar(const Lista &lista1, const Lista &lista2, Lista &listaResultado);

Hi, I think you've messed a little in the function suma () and it should be simpler, so I gave myself the task of altering your code a bit

void sumar(Lista &lista1, Lista &lista2, Lista &listaResultado){    
Lista l1=lista1, l2=lista2;

while (l1 != NULL && l2 != NULL) {

    int suma = l1->numero + l2->numero;
    insertarFinal(listaResultado, suma);
    l1 = l1->sig;
    l2 = l2->sig;
  }
}

Main

int main() {

    Lista lista1 = NULL, lista2 = NULL,listaSuma=NULL;
    int n;

   //Llenamos primer lista
    for (int i = 0; i < 3; i++) {
       cout << "Ingrese el valor lista 1 " << endl;
       cin >> n;
       insertarFinal(lista1, n);
    }
    cout << endl;

    //Llenamos segunda lista
    for (int i = 0; i < 3; i++) {
       cout << "Ingrese el valor lista 2" << endl;
       cin >> n;
       insertarFinal(lista2, n);
    }
    cout << endl;

    sumar(lista1, lista2, listaSuma);
    mostrar(listaSuma);

   cin.get();
   cin.get();
   return 0;
}

PS: Do not forget to free the memory.

basically you create a list 3 and insert the nodes of both lists with the primitives of inserting list in the third list