Exterior product is a ring, showing that the multiplication has desired qualities The most important Question for me is whether it is necessary or not to show that the functions are well defined
The book is Bosch Linear Algebra 1 page 159
Let $V$ be a Vector space we have showed that there exists a specific (i.e in the following discussion we use one particular) vectorspace (that we have defined and proved in beforehand and that has the following charachterizing charasterics, when we refer to ) $\bigwedge^rV$ (then we mean a certain canonical vectorspace) such that there exists a alternating  multilinear map (see below Definition bilinear) $\sigma:V^{r}\rightarrow\bigwedge^{r}V$ such that for every alternating multilinear map $\phi:V^{r}\rightarrow W$ where $W$ is anothere vectorspace there exists an unique linear map $\psi$ such that $\phi=\psi\circ\sigma$.
We have also proved that $\sigma(V^{r})=\sigma(a_1,...,a_r):=a_1\wedge...\wedge a_r$ with $a_1,...,a_r\in V$ is a generating System for $\bigwedge^{r}V$.

Now we want to Show that there is a  map $\wedge$ which is bilinear
and which fullfils the following properties for $r,s\neq 0$ 
$\wedge:\bigwedge^{r}V\times \bigwedge^{s}V\rightarrow
 \bigwedge^{r+s}V$.
$\wedge(a,b)=\wedge(a_1\wedge .... \wedge a_r,b_1\wedge .... \wedge
 b_s)=a_1\wedge...\wedge a_r\wedge b_1 \wedge....\wedge b_s$
we also want to show that there exists exactly one map with the
  properties above

A map $f:V\times V'\rightarrow W$ is bilinear if $f(\lambda a+\mu c,b)=\lambda f(a,b)+\mu f(c,b), f(a,\lambda b+\mu c)=\lambda f(a,b)+\mu f(a,c)$. We would call it also alternating if $V=V'$ and $f(a,a)=0$.
I have Problems to understand the proof
I will write down the proof and then talk about the parts that I have not understood, while I am writing the proof down I also make some comments on the parts that I don not understand. The main Purpose of those comments is to Show where I had Trouble.
First of all the map 
$$\phi:V^{r+s}\rightarrow \bigwedge^{r+s} V $$ 
$$(a_1,...,a_r,b_1,...,b_s)=a_1\wedge...\wedge a_r\wedge b_1\wedge b_s$$ 
is multilinear and alternating because it is the same canonical map we used in our definition.
We know that if we fix Elements $a_1,...,a_r$ then the map
$$V^{s}\rightarrow \bigwedge^{r+s} V,$$
$$a:(b_1,...,b_s)\mapsto a_1\wedge ... \wedge a_r \wedge b_1 \wedge .... \wedge b_s$$ is multilinear and alternating. Therefore we know there exists a linear map $b$ such that $a=b\circ \sigma_s$ (where $\sigma_s$ refers top the canonical bilinear map from $V^s$ to $\bigwedge^{s}V$ )   (understood)
Thus $\phi$ induces (don't know what this means, does it mean implies? Browsing only said that it has to do something with induction but how is induction applied here?) a map 
$$\phi':V^{r}\times \wedge^{s}V\rightarrow \bigwedge^{r+s}V$$
$$\big(a_1,...,a_r,\sum_jb_{j1}\wedge...\wedge b_{js}\big)\mapsto\sum_ja_1\wedge...\wedge a_r\wedge b_{j1}\wedge...\wedge b_{js}$$
and because Elements in $\bigwedge^{r+s}V$ that are written in the form $a_1\wedge....\wedge a_r\wedge b_1\wedge ...\wedge b_s$ are multilinear and alternating in each factor (how can an element be multiliear and alternating? Aren't those properties for certain function between vectorspaces? What could be meant with an element is alternating and multilinear in each of ist factor?) $\phi'(\cdot,b)$ for fixed $b\in \bigwedge^s V$ is multilinear and alternating on $V^{r}$. 
(in the previous Paragraph we wanted to Show that for a fixed element $b$ in $\bigwedge^sV$) the map $V^{r}\rightarrow \bigwedge^{r+s}(V)$ with $(a_1,....,a_r)\mapsto \phi'((a_1,...,a_r),b)$ is multilinear and alternating. I don't understand the $j$ in the index of the sum in $\phi'$ if we fix one element $b$ then we can write it as $b_1\wedge....\wedge b_s$ wherefore there is a $j$ in the sum what does it mean? Could it mean that it doesn't matter how $b$ is representated? This evokes a General Question, namely well-definition when or how do we prove that the representation of the Elements are not important?)
Coming back to the proof, the last observatuion Shows that $\phi'$ induces a function $\phi''$ with:
$$\phi'':\bigwedge^{r}V\times \bigwedge^{s}V\rightarrow \bigwedge^{r+s}V$$
$$\big(\sum_ia_{i1}\wedge ...\wedge a_{ir},\sum_{j}b_{j1}\wedge...\wedge b_{js}\big)\mapsto\sum_{ij}a_{i1}\wedge...a_{ir}\wedge b_{j1}\wedge...\wedge b_{js}$$
where $\phi''(\cdot,b)$ is linear for all $b\in\bigwedge^{s}V$ on $\bigwedge^r V$ Because of some calculationrules 
(for all Elements $a$ in $\bigwedge^{x}V$ with $a=a_1\wedge ... \wedge a_x$ we have $a_1\wedge ... \wedge (\alpha_i+\beta b_i)\wedge  ... \wedge a_x= \alpha a_1\wedge ... a_i ... \wedge a_x + \beta a_1\wedge ... b_i ...\wedge a_x$ and $a=0$ if there exists one component that is equal to another component)
we can show that $\phi''$ is bilinear and that it fullfils the condiction of $\wedge$. There exists only one such map because we have already proved in advance (see intro) that Elements of the type $a_1\wedge ... \wedge a_r, b_1\wedge ...\wedge b_s$ are a generating System for the vectorspaces $\bigwedge V^{r},\bigwedge^{s}V$ respectively. $\square$
Questions About the proof
How does $\phi$ induce $\phi'$, and $\phi'$ $\phi''$
How is the well-defintion proved of $\wedge$
What does the index $i,j$ mean in $\phi'$ i.e. $\phi''$ 
How can an element be multlinear and alternating in each of its factors (regarding $a_1\wedge .... \wedge a_r$)
 A: 
Thus $\phi$ induces (don't know what this means, does it mean implies? Browsing only said that it has to do something with induction but how is induction applied here?)

In this case, inducing some other map has nothing to do with induction. Rather, we also say that a map $\phi$ induces another map $\phi'$ if we obtain $\phi'$ from $\phi$ by using a universal property. 
A universal property is e.g. the defining property of the exterior power: For every alternating $n$-linear map $f\colon V^n\to W$ there exists a linear map $f'\colon \bigwedge^n V\to W$ such that $f=f'\circ\sigma_n$. We say that $f$ induces the map $f'$.
For your proof, this means that once you have shown that $\phi_b\colon V^s \to\bigwedge^{r+s} V$ is alternating multilinear, you obtain a map $\phi'_r\colon \bigwedge^s V\to\bigwedge^{r+s} V$. Then letting $\phi'(a, b):=\phi'_b(a)$ gives you a map $$\phi'\colon V^r\times\bigwedge^s V\to \bigwedge^{r+s} V.$$

[...] because Elements in $\bigwedge^{r+s}V$ that are written in the form $a_1\wedge....\wedge a_r\wedge b_1\wedge...\wedge b_s$ are multilinear and alternating in each factor (how can an element be multiliear and alternating? Aren't those properties for certain function between vectorspaces? What could be meant with an element is alternating and multilinear in each of ist factor?) $\phi'(\cdot,b)$ for fixed $b\in\wedge^s V$ is multilinear and alternating on $V^r$.

Your concern is right; it does not make sense to state that an element of $\bigwedge^{r+s}$ is alternating multilinear. But you have already used above that $\phi'_b=\phi'(\cdot, b)$ is alternating multilinear, so there is nothing left to do about this.

I don't understand the $j$ in the index of the sum.

The issue is that $\bigwedge^s V$ is a vector space generated by elements of the form $b_1\wedge\cdots\wedge b_s$. It is not true that any element $v$ of $\bigwedge^s V$ can be written as $v=b_1\wedge\cdots\wedge b_s$ for certain $b$'s. Rather, $v$ always is a linear combination of elements of that form. 
In other words any $v\in\bigwedge^s V$ can be written as $v=\sum_{j=1}^n b_{j1}\wedge\ldots\wedge b_{js}$, where all $b_{jk}\in V$.
Even more specific, if $V$ is finite dimensional with basis $b_1,\dots,b_n$, then the elements $b_{i_1}\wedge\ldots\wedge b_{i_s}$ for $i_1<i_2<\dots<i_s$ form a basis for $\bigwedge^s V$.

How does $\phi$ induce $\phi''$

The same way as it induces $\phi'$: You notice that the map $\phi'\colon V^r\times\bigwedge^s V\to\bigwedge^{r+s}V$ is alternating multilinear in its first argument; so it induces (see above) a map 
$$\phi''\colon \bigwedge^r V\times\bigwedge^s V\to \bigwedge^{r+s} V.$$

How is the well-defintion proved

For well-definedness, you plugin different expressions for the same element and verify that the output is unchanged. Hint: The equivalence relation involved in $\bigwedge^s V$ is that $b_1\wedge\dots\wedge b_s=\operatorname{sgn}(\sigma) b_{\sigma(1)}\wedge\dots\wedge b_{\sigma(s)}$. So you have to check that 
$$\phi(a_1\wedge\dots\wedge a_r, b_1\wedge\dots\wedge b_s) = \operatorname{sgn}(\sigma)\phi(a_1\wedge\dots\wedge a_r, b_{\sigma(1)}\wedge\dots\wedge b_{\sigma(s)}),$$
and the same for $a$. But this is immediate because you obtain this map from the defining properties of $\bigwedge^s V$.
Edit: Do you have a clue why this map is unique?
