(Assuming $f$ was meant to be generically finite)

Since the blowups are proper and $f$ is proper, the "property P argument" shows that $\tilde f$ is proper. A proper quasi-finite morphism is finite (EGA IV 18.12.4), so $\tilde f$ is finite.

This (more or less) reduces to the case when $f$ is finite to begin with, so no blowups are needed. I can't think of a nontrivial condition that guarantees that a finite morphism is flat.

Since the blowups are proper and $f$ is proper, the "property P argument" shows that $\tilde f$ is proper. A proper quasi-finite morphism is finite (EGA IV 18.12.4), so $\tilde f$ is finite.

This (more or less) reduces to the case when $f$ is finite to begin with, so no blowups are needed. Then the only condition I know to ensure flatness is the one Dave Anderson cited, [Matsumura's Commutative Ring Theory, Theorem 23.1]: if $Y$ is regular and $X$ is Cohen-Macaulay, then $f$ is flat.