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?
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.
