Inverse image of a line bundle on $\mathbb{P}^1$ and Euler-like exact sequence Let $E=\mathcal{O}_{\mathbb{P}^1}(-1)$. Then we have the following exact sequence
$$0\rightarrow E\rightarrow\mathcal{O}_{\mathbb{P}^1}^{\oplus 2}\rightarrow E^{-1}\rightarrow0.$$
This sequence can be rewritten as
$$0\rightarrow E\rightarrow\mathcal{O}_{\mathbb{P}^1}^{\oplus 2}\rightarrow \omega_{\mathbb{P}^1}^{-1}\otimes E\rightarrow0.\,\,\,\,(1)$$
Now let $X$ be a projective variety and $f:X\rightarrow\mathbb{P}^1$ be a morphism. Is it true that for some class of "nice" morphisms we will have the exact sequence which looks like
$$0\rightarrow f^*E\rightarrow\mathcal{O}_{X}^{\oplus 2}\rightarrow \omega_X^{-1}\otimes_{\mathcal{O}_X}f^*E\rightarrow0\,?$$
I don't know what the assumption on $X$ and $f$ should be. For instance, the interesting example for me is when $\pi:X\rightarrow\mathbb{P}^1\times\mathbb{P}^n$ is the blowing-up and $f=p\circ\pi$, where $p:\mathbb{P}^1\times\mathbb{P}^n\rightarrow\mathbb{P}^1$ is the projection onto the first component. 
 A: Yes, morphisms $f:X\to Y$ for which the functor $f^\ast: QCoh_Y\to QCoh_X$ is left exact are called flat. 
But I have a comment on the "right part" of the sequences you write: be careful that the pullback of the canonical sheaf is not the canonical sheaf in general, so the cokernel of your last short exact sequence may be something else. In general, for a flat (even locally complete intersection) morphism $f$ as above, you have
$$\omega_f=\omega_X\otimes f^\ast\omega_Y^{-1}.$$ Thus $f^\ast(\omega_{\mathbb P^1}^{-1}\otimes E)=f^\ast(\mathscr O_{\mathbb P^1}(2)\otimes \mathscr O_{\mathbb P^1}(-1))=f^\ast\mathscr O_{\mathbb P^1}(1)$. This, I think, should be the cokernel of the third short exact sequence: notice that, in fact, it coincides with $f^\ast(E^{-1})$.
Now, for your example, I do not have any useful comment I am afraid. I can only remind you that blow-ups are never flat, and that morphisms to nonsingular curves (like $\mathbb P^1$) are flat if and only if they are surjective. Hope this helps!
Edit. As for my statement on maps to curves, I was a bit sloppy. Let $C$ be a smooth (connected) curve. Then:


*

*A flat morphism $X\to C$ which is a closed map is surjective (the image being open and closed);

*If $f:X\to C$ is surjective and $X$ is reduced, then $f$ is flat. Indeed, for any $x\in f^{-1}(c)$, the map $\mathscr O_{C,c}\to \mathscr O_{X,x}$ makes $\mathscr O_{X,x}$ into a flat (equivalently: torsion-free, as $\mathscr O_{C,c}$ is a Noetherian domain) $\mathscr O_{C,c}$-module if and only if $$\textrm{Ass}_{\mathscr O_{C,c}}(\mathscr O_{X,x})=\{(0)\}.$$ (Cf. e.g. Wikipedia for this). This means $X$ has no embedded components at $x$, which is certainly true if $X$ is reduced. Maybe some expert can come up with more general statements...

