I have a problem with the geometric product:
In my book the unit trivector is defined like this: $(e_{1}e_{2})e_{3}=e_{1}e_{2}e_{3}$ But that would mean $(e_{1}e_{2})e_{3}= (e_{1} \wedge e_{2})\cdot e_{3}+(e_{1} \wedge e_{2} \wedge e_{3})$ But I thougt it is just $e_{1} \wedge e_{2} \wedge e_{3}$? I could somehow imagine in my head that the plane spanned by $e_{1} \wedge e_{2}$ is perpendicular to the line $e_{3}$ but I'm not sure that it works like that. Is that right? Would make sense. Ok but why is this true $(e_{1}\wedge e_{2})e_{1}=(-e_{2}e_{1})e_{1}=-e_{2}e_{1}e_{1}=-e_{2}$ because $(e_{1}\wedge e_{2})e_{1}=(e_{1} \wedge e_{2})\cdot e_{1}+ e_{1}\wedge e_{2} \wedge e_{1}$ where the last term is zero. I don't know how the term $e_{1} \wedge e_{2}$ interacts as dot product with $e_{1}$. Similar is $(e_{1}\wedge e_{2})(e_{2}\wedge e_{3})=e_{1}e_{3}$ why isn't this just zero? Because $(e_{1}\wedge e_{2})(e_{2} \wedge e_{3})=(e_{1}\wedge e_{2}) \cdot (e_{2} \wedge e_{3})+e_{1}\wedge e_{2} \wedge e_{2} \wedge e_{3}$ any help?
I think it's often easier to treat the dot and wedge products as shorthands for grade projection of geometric products.
For example, for vectors $a, b, c$:
$$a \cdot b = \langle ab \rangle_1, \quad a \wedge b = \langle ab \rangle_2$$
Similarly,
$$a \wedge b \wedge c = \langle abc \rangle_3$$
With that in mind, your book defines the unit trivector as $(e_1 e_2) e_3 = e_1 e_2 e_3$. The geometric product is associative, so that clearly must be true. Breaking it down into grades, however, gives this:
$$\begin{align*}e_1 e_2 e_3 &= \langle e_1 e_2 e_3 \rangle_1 + \langle e_1 e_2 e_3 \rangle_3 \\ &= (e_1 \wedge e_2) \cdot e_3 + (e_1 \cdot e_2) e_3 + e_1 \wedge e_2 \wedge e_3\end{align*}$$
You already realized that the first term is zero. The second term is clearly zero. So $e_1 e_2 e_3 = e_1 \wedge e_2 \wedge e_3$.
Because $e_1, e_2, e_3$ are all orthogonal, most geometric products between them reduce to wedge products and vice versa. Consider:
$$e_1 e_2 = \langle e_1 e_2 \rangle_0 + \langle e_1 e_2 \rangle_2 = e_1 \cdot e_2 + e_1 \wedge e_2 = e_1 \wedge e_2$$
Since the dot product orthogonal vectors is zero.
Hence, when you try to evaluate $(e_1 \wedge e_2) \cdot e_1$, you get
$$(e_1 \wedge e_2) \cdot e_1 = (e_1 e_2) \cdot e_1 = \langle e_1 e_2 e_1 \rangle_1 = -\langle e_2 e_1 e_1 \rangle_1$$
Group the $(e_1 e_1)$ together as $+1$, and you're done. The result is $-e_2$.
Part of your problem is that you're using $ab = a \cdot b + a \wedge b$ for everything, even when $a, b$ are not vectors. That's not correct practice.
Two general multivectors $A, B$ of grades $p, q$ have their geometric product decompose as follows:
$$AB = \langle AB \rangle_{p+q} + \langle AB \rangle_{p+q-2} + \ldots + \langle AB \rangle_{|p-q|}$$
For any term where the grade is greater than the dimension of the space, the grade projection is zero.
Hence, for two bivectors $p=q=2$, the terms are
$$AB = \langle AB \rangle_4 + \langle AB \rangle_2 + \langle AB \rangle_0$$
The grade-4 term is zero in 3d space, but it would be denoted with the wedge product. The grade-2 term is typically denoted $A \times B$ and called the commutator product. The grade-0 term is typically denoted with a dot product.
So when you multiply $(e_1 \wedge e_2)(e_2 \wedge e_3)$, the easy thing to do is to eliminate the wedges in favor of geometric products, and then compute the terms.
$$(e_1 \wedge e_2)(e_2 \wedge e_3) = e_1e_2 e_2 e_3$$
Group those $e_2 e_2$ together as $+1$, and you're done. The result is $+e_1 e_3$. This is the commutator product term that you didn't realize was there.
The main convenience of using an orthogonal basis is that you can do a lot of arithmetic directly with the geometric product acting on basis vectors. Basis vectors can be swapped at the cost of just a minus sign, and you can reduce the overall product until there are no more contractions of vectors to perform.