Let $\varphi:A \rightarrow B$ be a ring morphism. Consider the corresponding map of schemes, $f: Spec\ B \rightarrow Spec\ A$. This map $f$ is given by $\varphi ^{-1},$ since prime ideals pull back to prime ideals. Let's denote $X=Spec\ B$ and $Y=Spec\ A$ so $f:X \rightarrow Y.$ We also have a map of sheaves $ f^\# : \mathcal{O}_Y \rightarrow f_*\mathcal{O}_X, $ which by adjointness is equivalent to $ f_\# :f^* \mathcal{O}_Y \rightarrow \mathcal{O}_X. $ I know that the map of sheaves (actually I mean the map on global sections of the sheaves) is supposed to be $\varphi:A \rightarrow B,$ and I know $\mathcal{O}_X=B$ and $\mathcal{O}_Y = A.$ So what exactly are $ f^\#, f_\#, f_*\mathcal{O}_X, $ and $f^*\mathcal{O}_Y$ ?
I've looked at the definitions for direct image and inverse image sheaf, and neither seem very helpful. For example, $f_*\mathcal{O}_X$ is given by the following: for every open subset $U\subset Y, \ f_* \mathcal{O}_X (U):=\mathcal{O}_X(f^{-1}(U)) $ . But I said $f=\varphi^{-1}$ so then $f^{-1}=\varphi,$ which doesn't make sense because prime ideals do not always map to prime ideals.
It doesn't make sense to say that $f^{-1} = \varphi$. Remember, $\varphi$ is a function that takes an element of $A$ and gives us an element of $B$, whereas $\varphi^{-1}$ is a function that takes a subset of $B$ and gives us a subset of $A$; in particular, it is not the inverse function of $\varphi$ (which probably doesn't even exist, since $\varphi$ is usually not going to be an isomorphism).
Go back and look at that carefully and you'll see there's no problem there.
In general it's going to be pretty messy to describe what these maps look like for arbitrary choices of $U$, since an arbitrary open set $U$ is just the complement of some arbitrary collection of closed sets. However, we know that the open sets $D_f$ form a basis for the topology, and it's not too hard to describe what the maps
$$f^\#_{D_f} : \mathcal{O}_Y(D_f) \to (f_\ast \mathcal{O}_X)(D_f)$$
look like. This is surely done in your text, for instance.