Let $E$ be a normed $\mathbb R$-vector space.
On page 59 of Tensor Norms and Operator Ideals the authors define that $E$ has the approximation property, if for every absolutely convex compact subset $K$ of $E$ and $\epsilon>0$ there is a bounded linear operator $T:E\to E$ of finite rank with $\left\|Tx-x\right\|\le\epsilon$ for all $x\in K$.
On the other hand, on page 72 of *Introduction to Tensor Products of Banach Spaces, the author defines that a complete $E$ (i.e. a $\mathbb R$-Banach space $E$) has the approximation property, if the condition above is satisfied for every compact subset $K$ of $E$. Hence, he doesn't require $K$ to be absolutely convex.
I have almost no knowledge in this topic. I just want to understand, if both definitions are equivalent in the case of a $\mathbb R$-Banach space $E$. Are they?
The two properties are equivalent in Banach spaces. Clearly the second implies the first, since the family of absolutely convex compact sets in $E$ is contained in the family of compact sets.
Let's call $M \subset E$ "good" if for every $\epsilon > 0$ there is a continuous finite-rank operator $T \colon E \to E$ with $\lVert Tx - x\rVert \leqslant \epsilon$ for all $x\in M$. It is straightforward that every subset of a good set is good, so to see that the first property implies the second in Banach spaces it suffices to see that every compact subset of a Banach space is contained in an absolutely convex compact set.
The absolutely convex hull of a totally bounded set is totally bounded in locally convex spaces (for the convex hull, this is part (b) of theorem 3.20 in Rudin's Functional Analysis, the modification for the absolutely convex hull is easy), and hence the closure of the absolutely convex hull of a compact set is an absolutely convex compact set in Banach spaces.