Yes, even with no assumptions on $f$. Suppose we have an exact sequence
$$0\to M_n\to\cdots\to M_1\to M_0\to0$$
of flat $A$-modules. If $N$ is any $A$-module, then the sequence
$$0\to M_n\otimes_AN\to\cdots\to M_1\otimes_AN\to M_0\otimes_AN\to0$$
is exact. In the case of $f:\mathop{\mathrm{Spec}}B\to\mathop{\mathrm{Spec}}A$, pullback is given by $-\otimes_AB$, so this applies, and the general case reduces to the affine case.
To prove the claim, break the long exact sequence up into short exact sequences. Starting on the right we have
$$0\to K_1\to M_1\to M_0\to0$$
where $K_1=\ker(M_1\to M_0)=\mathrm{coker}(M_3\to M_2)$. Then we get a long exact sequence
$$\mathrm{Tor}_1^A(M_0,N)\to K_1\otimes_AN\to M_1\otimes_AN\to M_0\otimes_AN\to0.$$
But $M_0$ is flat, so the Tor vanishes, giving
$$0\to K_1\otimes_AN\to M_1\otimes_AN\to M_0\otimes_AN\to0.$$
Because $M_1$ is flat as well, the long exact sequence also shows that $\mathrm{Tor}_1^A(K_1,N)=0$, meaning $K_1$ is also flat. Now look at the next short exact sequence
$$0\to K_2\to M_2\to K_1\to0$$
where $K_2=\ker(M_2\to M_1)=\mathrm{coker}(M_4\to M_3)$. We have seen that $K_1$ is flat, so the same argument as above shows that
$$0\to K_2\otimes_AN\to M_2\otimes_AN\to K_1\otimes_AN\to0$$
is exact. And so on.
For $n>1$, the natural guess would be that $\pi_*\mathcal{O}_{\widetilde{X}}(-nE)=I_{Y/X}^n$, and it's true in this case. For $n<0$, we have $\pi_*\mathcal{O}_{\widetilde{X}}(-nE)=\mathcal{O}_X$.
To show these claims, we'll work locally on $X$. Assume $X=\operatorname{Spec} A$ is affine and $Y$ is cut out by the ideal $I_Y=(f_1,\cdots,f_r)$. Then we get a surjection $A^r\to (f_1,\cdots,f_r)$ which turns in to a surjection of the graded algebras $\operatorname{Sym}(A^r)\to \bigoplus_{m\geq 0} I_Y^m$ corresponding to the closed immersion $\widetilde{X}\hookrightarrow \Bbb P(A^r)$. Here, the exceptional divisor $E$ corresponds to the line bundle $\mathcal{O}_{\Bbb P(A^r)}(-1)|_{\widetilde{X}}$.
Next, via the restriction map $\mathcal{O}_{\Bbb P(A^r)}(m) \to \mathcal{O}_{\Bbb P(A^r)}(m)|_{\widetilde{X}}$ and the canonical isomorphism $\mathcal{O}_{\widetilde{X}}(1)\cong \mathcal{O}_{\widetilde{X}}(-E)$, we see that if $\mathcal{O}_{\Bbb P(A^r)}(n)$ is globally generated, then $\pi_*\mathcal{O}_{\Bbb P(A^r)}(n)\to \pi_*\mathcal{O}_{\widetilde{X}}(-nE)$ will be surjective and thus an isomorphism of line bundles. By identifying $\pi_*\mathcal{O}_{\Bbb P(A^r)}(n) = \operatorname{Sym}^n(A^r)$ and $\pi_*\mathcal{O}_{\widetilde{X}}(-nE)=I^n$, we see that when the global generation condition is satisfied, then we have an isomorphism $\mathcal{O}_{\widetilde{X}}(-nE)\cong I_Y^n$.
By Serre vanishing, this is always the case for any $X,Y$ assuming that $n>>0$. In our case, the argument from the linked post shows that in fact $\mathcal{O}_{\Bbb P(A^r)}(1)$ is globally generated, so $\mathcal{O}_{\Bbb P(A^r)}(n)$ is globally generated for all $n>0$. So we get our claimed isomorphism $\pi_*\mathcal{O}_\widetilde{X}(-nE)\cong I^n_Y$.
For $n<0$, after tensoring the natural exact sequence $0\to \mathcal{O}_\widetilde{X}(nE) \to \mathcal{O}_X\to\mathcal{K} \to 0$ by $\mathcal{O}_\widetilde{X}(-nE)$ we get the sequence $0\to \mathcal{O}_\widetilde{X}\to \mathcal{O}_\widetilde{X}(-nE)\to \mathcal{K}(-nE)\to 0$ (here $\mathcal{K}$ is the structure sheaf of a thickening of $E$). If we prove that $\pi_*\mathcal{K}(-nE)$ has no global sections, then it's the zero sheaf as $X$ is affine, and this would imply $\mathcal{O}_X=\pi_*\mathcal{O}_\widetilde{X}\to \pi_*\mathcal{O}_\widetilde{X}(-nE)$ is an isomorphism.
As $\mathcal{O}_\widetilde{X}(-nE)\cong \mathcal{O}_\widetilde{X}(n)$, we see that it's a negative line bundle, and after restriction to any projective subvariety of $\Bbb P(A^r)$ it will still be a negative line bundle and therefore have no sections. Taking the fiber $\Bbb P(A^r)_y$ for $y\in Y$, we see that this is a projective variety, so there are no sections of $\mathcal{O}_\widetilde{X}(n)$ in the fiber direction along $E\to Y$. But any global section of $\pi_*\mathcal{K}(-nE)$ would come from such a global section of $\mathcal{O}_\widetilde{X}(n)$ because $\mathcal{O}_\widetilde{X}(n)\to \mathcal{K}(-nE)$ is surjective and would remain this way after restricting to $E$. So there cannot be any global sections and we've shown that $\pi_*\mathcal{O}_\widetilde{X}(-nE)=\mathcal{O}_X$ for $n<0$.
Best Answer
No. Let $X=\Bbb P^3$ and $Y=\{[1:0:0:0]\}$ with $F'=\mathcal{O}_{\Bbb P^3}^3$ and $L'=\mathcal{O}_{\Bbb P^3}(1)$, and let $f':F'\to L'$ be given by the matrix $\begin{pmatrix} x_1 & x_2 & x_3 \end{pmatrix}$. Then this map is surjective on $X\setminus Y$, and by letting $\pi^*F'=F$ and $\pi^*L'=L$, we get the desired counterexample.