I am currently studying the Magnus property:
Let $G$ be a group and $u, v \in G$. If the normal closures of $u$ and $v$ coincide, then $u$ is conjugate to $v$ or $v^{−1}$.
I was told that this property was named after W. Magnus who proved the Freiheitsatz and that free groups have this property in this text in pages 141-165. However, this text is in German, which I am not fluent in, and is very technical, using multiple lemma's and notation that I am not familiar with.
I do have a decent understanding of group theory and was hoping that some body could give some intuitive way of seeing how the Freiheitsatz implies free groups having the Magnus property?
The result you state about free groups is Theorem 4.11 of the standard text Combinatorial group theory by Magnus, Karrass and Solitar* (p261 of my copy). The book includes an English-language proof of this result, as well as a proof of the Freiheitssatz itself. The proof in the book is 4.5 pages long, and the idea is to apply "Magnus' method". This method was first used by Magnus to study one-relator groups, and indeed is the method used in the text you link too.
However, these proofs are painful. They use something called "staggered presentations", which was Magnus' original way of applying his method. A more up-to-date version of Magnus' method, due to Moldavanski, uses HNN-extensions**. You can find detailed descriptions of this method in the paper McCool, Schupp, On one relator groups and HNN extensions, J. Aust. Math. Soc. 16(02):249--256 (1973), and also in these notes of Andy Putman. Neither resource contains a proof of the result that you are after, but I guess you can try to turn the handle if you are enthusiastic enough. Sounds like a nice master's project actually! :-)
I should say a few words about Magnus' method. Any proof which uses Magnus' method must apply the Freiheitssatz somewhere (in particular, the proof of Theorem 4.11 in Magnus, Karrass and Solitar - the proof that you are interested in - uses the Freiheitssatz, as you were told). Magnus' method is an inductive argument on the length of the word $R$. An explicit example of a single step in the HNN-version of the method is as follows: Suppose $R=ab^2c^2a^{-1}b^2c^2$ and consider the group $G=\langle a, b, c\mid R\rangle$. We have the following: $$ \begin{align*} \langle a, b, c\mid ab^2c^2a^{-1}b^2c^2\rangle&\cong\langle a, b_0, c_0, b_1, c_1\mid ab_0^2c_0^2a^{-1}b_0^2c_0^2, ab_0a^{-1}=b_1, ac_0a^{-1}=c_1\rangle\\ &\cong\langle a, b_0, c_0, b_1, c_1\mid b_1^2c_1^2b_0^2c_0^2, ab_0a^{-1}=b_1, ac_0a^{-1}=c_1\rangle \end{align*} $$ By the Freiheitssatz, the subgroup $\langle b, c\rangle$ of $G$ is free of rank two. Hence, $\langle b_0, c_0\rangle$ and $\langle b_1, c_1\rangle$ are also free of rank two. Therefore, the above working shows that $G$ is an HNN-extension of the group $G_1=\langle b_0, c_0, b_1, c_1\mid b_1^2c_1^2b_0^2c_0^2\rangle$. This new group $G_1$ has a single defining relator which is shorter than the relator of the old group, so we have our induction step.
*This book is usually referred to as "Magnus, Karrass and Solitar" to distinguish it from "Lyndon and Schupp"; both sets of authors wrote books called "combinatorial group theory", and both books are still standard texts in geometric group theory.
**Jim Howie gave an alternative version of the method using "staggered complexes", while a third method used "pictures" (I cannot find out at the moment who first used this third method, but I know Hamish Short used it in his early work). All three methods are essentially equivalent, but HNN-extensions are much more standard in group theory and so the proofs are more readable to the generalist.