When you perform the substitution described in the preceding paragraph, you’re replacing the equations of these axes with the equations of their preimages. That is, every term that involves $x$ is in a sense talking about the $y$-axis ( $x=0$) and every term that involves $y$ is talking about the $x$ axis. Observe that the equations of the coordinate axes are hidden in the implicit equation $F(x,y)=0$. You still have the problem of constructing the reflection through an arbitrary line, but here’s an alternative viewpoint that might work without introducing too many new concepts.

A reflection is its own inverse, so in that special case the inversion just involves replacing $x$ with $x'$ and $y$ with $y'$ in the transformation equations. For an affine transformation, this is a simple matter of solving a pair of linear equations. It shouldn’t be too hard to get this general principle across: you essentially solve for $x$ and $y$ in the transformation equations and substitute into the equation of the curve. You know how if you have the line $y=x$, and you want to reflect the graph of a function $f(x)$ across it, you can just switch the $y$ and $x$ in the equation of the function (since you're just finding $f^)(x',y')=0$.

Here's a question one of my precalculus students asked me, paraphrased for clarity:

Student reflection graph