A map $B \longrightarrow A$ does not generally describe a homeomorphism $\operatorname{Spec} A \longrightarrow \operatorname{Spec} B$, but it does describe a continuous map. For instance, the map $\mathbb Z \longrightarrow \mathbb Z/2$ corresponds to the inclusion $\{(2)\} \subseteq \operatorname{Spec} \mathbb Z$ - certainly not a homeomorphism.
Also, you're right that a map of schemes requires a map of the structure sheaves. This is not derivable from the map on the underlying space. Instead it comes from the map of rings. Indeed, the idea is that for a basic open set $D(f) \subseteq \operatorname{Spec} B$, its pullback under the map induced by $\phi: B \longrightarrow A$ is going to be $D(\phi(f))$. Recall that $\Gamma(D(f), \operatorname{Spec} B) = B_f$. We're looking for a map $\Gamma(D(f), \operatorname{Spec} B) \longrightarrow \Gamma(D(\phi(f)), \operatorname{Spec} A)$ to define our map of structure sheaves. We'll then take this to be the map $B_f \longrightarrow A_{\phi(f)}$, which we get from the universal property of localization applied to $\phi$. Now, I've only defined the map on a basis of open sets, but general properties of sheaves imply that this is enough. Also, you need to check that the induced maps on stalks are all local, but I won't show this. Here is a reference to the Stacks project for this fact.
Now I hope this construction explains why a ring homomorphism induces a reverse map of affine schemes, structure sheaf and all. But I did say that you cannot derive this map purely from the map on the underlying topological spaces, and I'd like to explain that further. Indeed, in defining a map of affine schemes it would suffice to just specify a continuous map if, for instance, the forgetful functor $F: \mathbf{AffineSchemes} \longrightarrow \mathbf{Top}$ was fully faithful. In fact, it is neither full mor faithful.
For fullness, consider maps $\operatorname{Spec} \mathbb Z \longrightarrow \operatorname{Spec} \mathbb Z$. There is only one ring homomorphism $\mathbb Z \longrightarrow \mathbb Z$, so the only map between the affine schemes $\operatorname{Spec} \mathbb Z \longrightarrow \operatorname{Spec} \mathbb Z$ is the identity. However, there are many continuous maps between the underlying spaces $F(\operatorname{Spec} \mathbb Z) \longrightarrow F(\operatorname{Spec} \mathbb Z)$. For instance, there are infinitely many constant maps, all of which are continuous. Hence, the forgetful functor $F$ cannot be full. In other words, not every map of the underlying topological space can arise from a map of affine schemes.
Now for faithfulness, consider $\mathbb Q(i)$. We have two automorphisms of this field, the identity and complex conjugation. They therefore define two automorphisms of the affine scheme $\operatorname{Spec} \mathbb Q(i)$. However, $\mathbb Q(i)$ is a field so the topological space of the affine scheme, $F(\operatorname{Spec} \mathbb Q(i))$, is a single point. There is only one continuous map from a point to itself, so $F$ cannot be faithful. That is, it is impossible in general to take a continuous map between affine schemes and derive a corresponding map on the structure sheaves as the same continuous map can come from many scheme maps.
These examples also show that the forgetful functor $\mathbf{Schemes} \longrightarrow \mathbf{Top}$ is neither full nor faithful, so you cannot define a map of schemes purely from a continuous map. I'd also like to point out that unlike affine schemes, a map between general schemes is not determined by a (reversed direction) ring homomorphism on global sections. Indeed, consider projective space over an algebraically closed field $k$. The global sections of $\mathbb P_k^n$ is $k$ for all $n$. There are many maps between projective spaces that are constant on global sections. For instance, linear automorphisms of $\mathbb P_k^n$ are defined by matrices in $PGL_n(k)$, and in general this group is nontrivial.
Best Answer
You can find answers to both of these questions in Hartshorne's Algebraic Geometry.
Your first question is Proposition II.5.8(c). As a sketch of the proof, since $X$ is Noetherian, it can be covered by finitely many affine open sets $U_i$, and each intersection $U_i\cap U_j$ can also be covered by finitely many affine open sets $U_{ijk}$. Now, the pushforward of $\mathcal{F}$ restricted to any $U_i$ or $U_{ijk}$ is quasicoherent. But the pushforward of $\mathcal{F}$ is just the kernel of a certain morphism between products of these restrictions, by the sheaf gluing axiom (a section of $f_*\mathcal{F}$ over $U$ is the same as sections over $U\cap U_i$ for each $i$ such that the restrictions to the $U_{ijk}$ agree). Since the kernel of a morphism of quasicoherent sheaves is quasicoherent, this implies $f_*\mathcal{F}$ is quasicoherent. (The Noetherian hypothesis is used to guarantee that the products of the restricted sheaves are finite, since an infinite product of quasicoherent sheaves may not be quasicoherent.)
Your second question follows from the fact that cohomology of sheaves can be defined as derived functors of the functor of global sections on the category of sheaves of abelian groups on the space. The category of sheaves of abelian groups depends only on the topological structure (not the ringed space structure), and so this definition is obviously preserved by homeomorphisms. The equivalence of cohomology defined for sheaves of abelian groups and cohomology defined for sheaves of $O_X$-modules is Proposition III.2.6 in Hartshorne. As a sketch of the proof, you can show that flasque sheaves are acyclic in both categories, and so you can compute the cohomology of an $O_X$-module in either category by taking a resolution by flasque $O_X$-modules.