Of course, the choice of lattice you use is just an intermediate step in a calculation, and has no bearing on the final results. But a bad choice of lattice might make the result harder to see.
Suppose that the scattering for any kind of particle whatsoever responds to a potential $\rho(\mathbf{x})$. Under the Born approximation, the amplitude to scatter a particle and change its momentum by $\mathbf{k}$ is proportional to
$$\widetilde{\rho}(\mathbf{k}) = \int d\mathbf{x} \, e^{i \mathbf{k} \cdot \mathbf{x}} \rho(\mathbf{x}).$$
For phonons, Bloch's theorem tells us that
$$\rho(\mathbf{x}) = u(\mathbf{x}) e^{i \mathbf{k}' \cdot \mathbf{x}}$$
where $\mathbf{k}'$ is the phonon's crystal momentum, and $u(\mathbf{x})$ is periodic. Because of this periodicity, it's useful to do the integral in two stages. We decompose it into an integral over a "unit cell" $V$, and a sum over such cells. This gives
$$\widetilde{\rho}(\mathbf{k}) = \sum_{\text{cells at } \mathbf{x}_i} e^{i (\mathbf{k} + \mathbf{k}') \cdot \mathbf{x}_i} S(\mathbf{k}), \quad S(\mathbf{k}, \mathbf{k}') = \int_V d\mathbf{x} \, e^{i(\mathbf{k} + \mathbf{k}') \cdot \mathbf{x}} u(\mathbf{x})$$
where the integral $S(\mathbf{k}, \mathbf{k}')$ is called a form factor. Performing the sum,
$$\widetilde{\rho}(\mathbf{k}) \propto S(\mathbf{k}, \mathbf{k}') \sum_{\mathbf{G}} \delta(\mathbf{k} + \mathbf{k}' - \mathbf{G})$$
where the sum is over reciprocal lattice vectors $\mathbf{G}$.
When you change the choice of unit cell, you just change what goes into the form factor, and what goes into the sum, but the overall result is of course the same. In your first example, the scattering amplitude off the indicated optical phonons vanishes because the sum vanishes -- that is, whenever the frequencies match, none of the delta functions in the sum are nonzero. In the second example, it vanishes because the form factor vanishes: at the point you mark, it has the form $+1 - 1 = 0$. Both options are valid.
