If you have say an affine variety defined over $\mathbb{R}$, then its image under a morphism (also defined over $\mathbb{R}$) is a constructible set. But presumably there would be no good reason in general why the image of the set of $\mathbb{R}$-rational points under the morphism would have to be equal to the set of $\mathbb{R}$-rational points of the constructible set in question.


Or is there some useful sufficient condition for this to happen?



image under a morphism of a variety defined over R
You are right that in general this is not the case. For example let $X$ be the zero set of $x-y^2$ in $\mathbb{A}^2$ and consider the projection $X \to \mathbb{A}^1, (x,y) \mapsto x$.


Perhaps you are looking for something like that:
Let $f:X \to Y$ be a finite surjective morphism of real smooth varieties. Assume that the real points are dense in $X$, that the real points of $Y$ are connected and that $f$ is unramified over the real points of $X$. Then the image of the real points should be the real points of the image.


