Does the appliance of R-squared to non-linear models depends on how we calculate it?

Does the appliance of R-squared to non-linear models depends on how we calculate it? $R^2 = \frac{SS_{exp}}{SS_{tot}}$ is going to be an inadequate measure for non-linear models since an increase of $SS_{exp}$ doesn't necessarily mean that the variance is decreasing, but if we calculate it as $R^2 = 1 - \frac{SS_{res}}{SS_{tot}}$, then it's as much meaningful for non-linear models as it is for linear ones. I asked a similar question here where I showed that R-squared is no worse for non-linear models

So, what is the particular reason to say that in the case of non-linear models $\mathbf{R^2}$ loses its interpretation of proportion of variance explained? In both cases (I mean linear and non-linear models) we learn by how much your model's variance decreased with respect to its initial (total) variance. If $R^2 = 0.86 \text %$ then your model's variance decreased by $0.86 \text %$ (no matter whether it's linear or not).

EDIT:

1. $SS_{tot} = \|y - \bar y\|^2$
2. $SS_{exp} = \|\hat y - \bar y\|^2$
3. $SS_{res} = \|y - \hat y\|^2$.

Where:

1. $y$ is a vector of true answers;
2. $\bar y$ is a vector whose elements are mean of $y$;
3. $\hat y$ is a vector with our model's predictions.