A continuous map does not tear. A local homeomorphism also "locally does not glue". This is captured in part by the fact it has discrete fibers: the points which are glued are isolated from each other in the domain.
But there may be a point whose every neighborhood intersects some fiber more than once, e.g $x\mapsto x^2$ as a function $(-\infty,\infty)\to [0,\infty)$. This map creates the boundary point zero. Thus the condition $f(\partial A)\supset \partial f(A)$ of reflecting boundaries seems reasonable as well. It also seems that reflecting boundaries should be equivalent to being open, which leads me to my questions:
Is a continuous open function with discrete fibers necessarily a local homeomorphism? If not, what more is needed?
Given such a map $f:X\to Y$ and a point $x\in X$, I don't see how to produce an open neighborhood of it which does not intersect any fiber more than once. I am stuck even if I assume $X$ is Hausdorff, so I think my intuitive picture is missing something.
As the accepted answer points out, local injectivity is the key property here which precisely captured the points have a neighborhood not intersecting any fiber twice. This implies discreteness of fibers and so a function is a local homeomorphism iff it's continuous, open, and locally injective.
No. For instance, the map $z\mapsto z^2$ from $\mathbb{C}\to\mathbb{C}$ is continuous, open, and has discrete fibers, but is not a local homeomorphism. The only extra condition I can think of you might add is requiring the map to be locally injective as well. This of course then implies it is a local homeomorphism, since if $f:X\to Y$ is injective when restricted to an open set $U$, then $f(U)$ is open in $Y$ and $f|_U$ is an open continuous bijection $U\to f(U)$ and hence a homeomorphism.