Lemma Let $\phi:Y\to X$ be a proper birational morphism of complex manifolds of dimension at least $2$. Let $E\subset Y$ be the exceptional locus of $\phi$. Note that $E$ is a Cartier divisor. Then for any $a>0$, $$\phi_*\mathscr O_E(aE)=0.$$

Remark Actually the statement is true for more general $\phi$, but this will do for now.

Proof (Sketch) We prove this by induction. If $\dim X=2$, then $E^2<0$, so $\mathscr O_E(aE)$ has no global sections and the statement follows. If $\dim X>2$, then we first take hyperplane sections on $X$ until $\phi(E)$ is zero-dimensional and then take hyperplane sections on $Y$. In both cases it is relatively easy to prove that the induction hypothesis implies the desired statement (I will try to add this later when I will have more time). Q.E.D.

Corollary Under the same assumptions as above, $$\mathscr O_X\simeq \phi_*\mathscr O_Y(aE).$$

Proof Apply $\phi_*$ to the short exact sequence $$0\to \mathscr O_Y((a-1)E) \to \mathscr O_Y(aE) \to \mathscr O_E(aE) \to 0.$$ Q.E.D.

Now as $X$ is smooth, we have $$K_Y\sim \phi^*K_X + aE$$ for some $a> 0$. Then by the projection formula $$\phi_*\mathscr O_Y(K_Y)\simeq \mathscr O_K(K_X)\otimes \phi_*\mathscr O_Y(aE)\simeq \mathscr O_X(K_X).$$ By the definition of $\phi_*$ the statement follows.

Remark this is actually true under more general circumstances.

Lemma Let $\phi:Y\to X$ be a proper birational morphism of complex manifolds of dimension at least $2$. Let $E\subset Y$ be the exceptional locus of $\phi$. Note that $E$ is a Cartier divisor. Then for any $a>0$, $$\phi_*\mathscr O_E(aE)=0.$$

Remark Actually the statement is true for more general $\phi$, but this will do for now.

Proof (Sketch) We prove this by induction. If $\dim X=2$, then $E^2<0$, so $\mathscr O_E(aE)$ has no global sections and the statement follows. If $\dim X>2$, then we first take hyperplane sections on $X$ until $\phi(E)$ is zero-dimensional and then take hyperplane sections on $Y$. In both cases it is relatively easy to prove that the induction hypothesis implies the desired statement (I will try to add this later when I will have more time). Q.E.D.

Corollary Under the same assumptions as above, $$\mathscr O_X\simeq \phi_*\mathscr O_Y(aE).$$

Proof Apply $\phi_*$ to the short exact sequence $$0\to \mathscr O_Y((a-1)E) \to \mathscr O_Y(aE) \to \mathscr O_E(aE) \to 0.$$ Q.E.D.

Now as $X$ is smooth, we have $$K_Y\sim \phi^*K_X + aE$$ for some $a> 0$. Then by the projection formula $$\phi_*\mathscr O_Y(K_Y)\simeq \mathscr O_K(K_X)\otimes \phi_*\mathscr O_Y(aE)\simeq \mathscr O_X(K_X).$$ By the definition of $\phi_*$ the statement follows.

Remark this is actually true under more general circumstances.

I've just noticed that I have not addressed your last question about why you need $(n-1)E$ added. One way to see this is by explicit calculation. Write down a differential form and the blow up in local coordinates and you see that when you pull it back the differential form picks up some poles, in fact, $(n-1)$ times $E$ poles.

Another way to compute this is to use adjunction: $$\mathscr O_Y(K_Y+E)|_E=\mathscr O_E(K_E)= \mathscr O_E(-n)$$ (since $E\simeq \mathbb P^{n-1}.$) At the same time, you know that the normal bundle of $E$ is $\mathscr O_E(-1)$ which is equivalent to saying that $\mathscr O_E(E)\simeq \mathscr O_E(-1)$. Combining these and the fact that the pull back of a line bundle restricted to a fiber is always trivial gives you what you want.