EDIT: I want to add that the relevant parts of EGA to compare are [EGAI, Thm. 1.7.3], which is the analogue of [Hartshorne, II, Prop. 2.3(c)], and [EGAI, Prop. 2.2.4], which is the analogue of your exercise. This proof is similar to the other answer.
[EGAInew, Prop. 1.6.3] is what I am paraphrasing below. It is also [EGAII, Err$_\mathrm{I}$, Prop. 1.8.1], with attribution to Tate.
I won't write down all of the details, but here is another way to approach the problem, which I think is easier, since it avoids the issue with trying to cover $X$ by open affines and glueing morphisms together. We use that the category of schemes is a full subcategory of the category of locally ringed spaces. It suffices to show
\begin{align*}
\alpha\colon \operatorname{Hom}_\mathsf{LRS}(X,\operatorname{Spec} A) &\longrightarrow \operatorname{Hom}_\mathsf{Ring}(A,\Gamma(X,\mathcal{O}_X))\\
(f,f^\#) &\longmapsto f^\#(\operatorname{Spec} A)
\end{align*}
is bijective. We construct an inverse map
$$
\rho\colon \operatorname{Hom}_\mathsf{Ring}(A,\Gamma(X,\mathcal{O}_X)) \longrightarrow \operatorname{Hom}_\mathsf{LRS}(X,\operatorname{Spec} A)
$$
as follows. Let $\varphi\colon A \to \Gamma(X,\mathcal{O}_X)$ be given. Define
$$
f \colon X \to \operatorname{Spec} A, \quad x \mapsto \{s \in A \mid \varphi(s)_x \in \mathfrak{m}_x\}
$$
where $\varphi(s)_x$ is the image of $\varphi(s)$ in the stalk $\mathcal{O}_{x,X}$ and $\mathfrak{m}_x \subseteq \mathcal{O}_{x,X}$ is the maximal ideal of $\mathcal{O}_{x,X}$. Note the set on the right is a prime ideal. The map $f$ is continuous since $f^{-1}(D(r)) = \{x \in X \mid \varphi(r)_x \notin \mathfrak{m}_x\} = D(\varphi(r))$. We define the map $f^\#$ of structure sheaves; since $D(r)$ form a basis of $\operatorname{Spec} A$, we construct the morphism on each $D(r)$ and then glue. We define $f^\#(D(r))$ to be the top arrow in the diagram
$$
\require{AMScd}
\begin{CD}
A_r @>f^\#(D(r))>\exists!> \mathcal{O}_X(f^{-1}(D(r)))\\
@AAA @AAA\\
A @>\varphi>> \mathcal{O}_X(X)
\end{CD}
$$
induced by the the universal property of localization [Atiyah-Macdonald, Prop. 3.1], where the hypotheses for the universal property hold since $\varphi(r)$ is invertible in $\mathcal{O}_X(f^{-1}(D(r)))$ by definition of $f$. The morphisms on each $D(r)$ glue together since the maps $f^\#(D(r))$ were constructed uniquely by the universal property above, hence on any intersection $D(rs)$ they must match.
To show $\alpha$ and $\rho$ are inverse to each other, note $\alpha \circ \rho = \mathrm{id}$ is clear by letting $r = 1$ in the diagram above. This implies $\alpha$ is surjective, so it remains to show $\alpha$ is injective. Let $\varphi\colon A \to \Gamma(X,\mathcal{O}_X)$, and let $(f,f^\#)$ such that $\alpha(f,f^\#) = \varphi$. Then, we have the diagram
$$
\begin{CD}
A_{f(x)} @>f^\#_x>> \mathcal{O}_{x,X}\\
@AAA @AAA\\
A @>\varphi>> \mathcal{O}_X(X)
\end{CD}
$$
by taking the direct limit over all open sets $D(r)$ containing a point $x$. Since the map $f_x^\#$ is local, we have $(f_x^\#)^{-1}(\mathfrak{m}_x) = \mathfrak{m}_{f(x)}$, hence $f(x) = \{s \in A \mid \varphi(s)_x \in \mathfrak{m}_x\}$ as desired by using the commutativity of the diagram. The uniqueness of $f^\#$ also follows from this diagram since if $(g,g^\#)$ is any other map $X \to \operatorname{Spec}A$ such that $\alpha(g,g^\#) = \varphi$, then $f^\#_x = g^\#_x$ for all $x$, hence they must be the same morphism.
Q1: On an affine scheme $Z$, we know that for any quasicoherent sheaf $\mathcal{A}$ we have that $\mathcal{A}\cong \widetilde{\mathcal{A}(Z)}$. Applying that to the case at hand, we know that $f_*\mathcal{F}$ is a quasicoherent sheaf on $Y$, so $f_*\mathcal{F}\cong \widetilde{f_*\mathcal{F}(Y)}$. But by the definition of the pushforward, we have that $f_*\mathcal{F}(Y) = \mathcal{F}(f^{-1}(Y)) = \mathcal{F}(X)$, and so we may conclude that $f_*\mathcal{F} \cong \widetilde{\mathcal{F}(X)}$.
Q2: One functor described here is just $R^\bullet f_*(-)$, which is known to be a cohomological $\delta$-functor basically by definition (it's a right derived functor of a left exact functor). The other functor here is the functor $\widetilde{H^\bullet(X,-)}$, which can be written as the composite of the two functors $\widetilde{-}$ and $H^\bullet(X,-)$. The first functor, $H^\bullet(X,-)$ is already a $\delta$-functor, and $\widetilde{-}$ being exact implies that it preserves the exactness of all the sequences and diagrams required to verify $H^\bullet(X,-)$ as a $\delta$-functor, so the composite is a $\delta$-functor.
Q3: Yes, this is essentially correct.
Best Answer
It's not true in general that $\widetilde{\mathcal F(X)}=\mathcal F$; there exist $\mathcal O_X$-modules which are not of the form $\widetilde M$ for any $M$.
However, what we do know is that there is a restriction map $\mathcal F(X)\to\mathcal F(X_f)$ for any $f\in A$. But $\mathcal F(X_f)$ is a module over $\mathcal O_X(X_f)=A_f$, so multiplication by $f$ is an automorphism of $\mathcal F(X_f)$, so by the universal property of localization of modules there is an induced ($A_f$-module) morphism $\mathcal F(X)_f\to\mathcal F(X_f)$ (making the corresponding triangle commute).
Combining with the map $M_f\to \mathcal F(X)_f$ you found, we get a map $\widetilde M(X_f)=M_f\to\mathcal F(X_f)$, and these you can glue to get a sheaf morphism $\widetilde M\to\mathcal F$.