Suppose that $f\colon X\to Y$ is a flat morphism of varieties over an algebraically closed field $k$. Let $E\subseteq X$ and $F\subseteq Y$ be closed subvarieties such that $f(E) = F$. Is it true that the restricted morphism $f|_E\colon E\to F$ is also flat? If not, are there some additional conditions on $f$ which would make this true?
No, the restriction $f:E\to F$ needn't be flat.
Take $Y=\mathbb A^2, X=\mathbb A^2 \times \mathbb P^1$ and for $f:X\to Y $ take the first projection, which is flat.
Now inside $X$ lies the blow-up $B\subset X$ of $Y$ at the origin $(0,0)\in \mathbb A^2=Y$.
The restricted map $f\mid B: B\to \mathbb A^2$ is well known not to be flat, because all its fibers outside the origin are single points, whereas the fiber at the origin is the one-dimensional projective space $\mathbb P^1$: flat maps do not tolerate such dimension jumps.
As to your second question, I am pessimistic about a general criterion ensuring that the restriction of a flat map will remain flat.
Flatness is a subtle relation between the fibers of a morphism, and I have the feeling that restricting a flat morphism to a subvariety of its domain will usually destroy this relation.