How is Post's tautology theorem used in this proof? Could someone please explain to me how does the proof of I.4.3 reference I.4.1?
In I.4.3, you are given hypotheses about A and B being theorems. However, I.4.1 talks about tautologies (as inputs) not theorems.
How can someone use $\Gamma \vdash A$ + something about tautologies to yield $\Gamma \vdash B \implies A$?
See :Tourlakis (2003), Lectures in logic and set theory.

 A: We have to use I.3.7 Definition (Tautologically Implies), page 31 :

Let $\mathcal A$ and $\Gamma$ be respectively any formula and any set of formulas.
  The symbol $\Gamma \vDash_{\text{Taut}} \mathcal A$, pronounced “$\Gamma$ tautologically implies $\mathcal A$”, means that every truth assignment $v$ that satisfies $\Gamma$ also satisfies $\mathcal A$.

We have $\mathcal A \vDash_{\text{Taut}} \mathcal B \to \mathcal A$ as well as $\mathcal B \vDash_{\text{Taut}} \mathcal A \to \mathcal B$.
We have to use them in the proof of :

I.4.3 Metatheorem. Any two theorems $\mathcal A$ and $\mathcal B$ are provably equivalent (page 42).

Assume $\Gamma \vdash \mathcal A$. 
Using I.4.1 Metatheorem (Post’s “Extended” Tautology Theorem), we have $\mathcal A \vdash \mathcal B \to \mathcal A$.
Thus, by properties of $\vdash$, we get : $\Gamma \vdash \mathcal B \to \mathcal A$.
In the same way, from $\Gamma \vdash \mathcal B$ we have : $\Gamma \vdash \mathcal A \to \mathcal B$.
Finally, using : $\mathcal A \to \mathcal B,  \mathcal B \to \mathcal A \vDash_{\text{Taut}} \mathcal A \leftrightarrow \mathcal B$, we conclude with :


$\Gamma \vdash \mathcal A \leftrightarrow \mathcal B$.


A: In Hilbert style proof systems, it is usually an axiom that the schema $\phi\to(\psi\to \phi)$ is tautological.   Additionally, Modus Ponens is a fundamental rule of inference (typically the only fundamental rule). 
So since $A\to(B\to A)$ is a tautology, and using Modus Ponens, therefore if can you derive $A$, then you may infer that $B\to A$ is derivable in the same context.$$\begin{split}\Gamma &\vdash A&&\textsf{derived somehow}\\&\vdash A\to(B\to A)&\qquad&\textsf{via axiom P2}\\\hline\Gamma &\vdash B\to A&&\textsf{via modus ponens} \end{split}$$
