Let $X$ and $Y$ denote topological spaces and $f : X \rightarrow Y$ denote a continuous mapping. Then $f$ is an embedding iff the corestriction $X \rightarrow f(X)$ is a homeomorphism. However this is quite a heavily packaged definition and I can imagine that it would sometimes be desirable to have a more direct characterization. I've thought about it a bit and I think that $f$ is an embedding iff firstly it is injective, and secondly we have that for all $U \subseteq Y$, if $f^{-1}(U)$ is open, then there exists an open set $V \subseteq Y$ such that $U \cap \mathrm{img}(f) = V \cap \mathrm{img}(f).$
Question. Is this a correct 'unpacking' of this notion, and if not, can anyone suggest a correct one?
Your unraveling is correct.
$f$ is an embedding in category $\mathbf{Top}$ if $f$ is injective and secondly $X$ is equipped with the coarsest topology that makes $f$ continuous.
The second statement comes to the same as: $\tau_X=\{f^{-1}(V)\mid V\in\tau_Y\}$. So for every $W\in\tau_X$ a set $V\in\tau_Y$ exists with $W=f^{-1}(V)$.
So if you start with some $U\subseteq Y$ and it appears that $f^{-1}(U)\in\tau_X$ then some $V\in\tau_Y$ exists with $f^{-1}(U)=f^{-1}(V)$ or equivalently $U\cap\mathsf{im}f=V\cap\mathsf{im}f$.
This might interest you as well.
In the category $\mathbf{Top}$ a continuous mapping $f:X\to Y$ is an embedding if it is injective and initial.
The continuous mapping $f$ is initial if for every topological space $Z$ and every function $g:Z\to X$ continuity of $f\circ g$ implies continuity of $g$.