Let $X$ be a scheme finite type over a field. The embedded points of $X$ are defined as the embedded points of $\mathcal{O}_X$ as $\mathcal{O}_X$-module.
I'm wondering that is there a more geometric way to understand embedded points? For example, what are embedded points look like when $\dim X=1$?
If $X$ has no embedded point, then one can show that every irreducible component of $X$ is one-dimensional. I'm wondering if the converse holds, i.e. if every irreducible component of $X$ has dimension one, then $X$ has no embedded point?
Embedded points care about things that are more scheme-theoretic in nature - for instance, $\operatorname{Spec} k[x,y]/(xy,y^2)$ has an embedded point $(x,y)$ but the underlying topological space is the same as $\Bbb A^1$. So this already shows that your final question has a negative answer.
For a generically reduced 1-dimensional scheme of finite type over a field, embedded points are precisely the non-reduced closed points. Omitting the generically reduced assumption, things get more complicated, but in general, one should think of embedded points as generic points of subvarieties that have more nilpotents than their surroundings.