Systems of linear equations: What did the author mean? EDIT: Since even after putting up a bounty I didn't get as much as a comment in two days I completely rewrote my question in hope to make it more accesible and get some good answers.
Consider the system in echelon form (the fat "X" denotes some sums):

This gives rise to a map $$\phi:\mathbb{R}^s\rightarrow L\subseteq \mathbb{R}^n,$$ where $s$ is the number of free variables for this system, that assign to every value of the free variable a solution ($L$ is the solution set). 
The author of lecture notes from where I copied the above continues saying, that this map is surjectiv, "which can be proved directly, but the proof is messy. We can show it in a more elegant fashion using the theory which we will learn later, since in particular we have to show that the number $r$ does not depend on our way of arriving at the above system, but only on the original system".
Up to the point of introducing this system, the concepts of "vector space", "rank" and "subspace" were not yet discussed. It was only showed that 
- we can bring every arbitrary system via row operations to a system of the above form
- that row transformations don't change the solution.
- that we can solve the above system by substituting backwards (if all the $b_{r+1},\ldots,b_n$ are $0$).
My question is: Why does the proof that $\phi$ is surjective require "later theory" ? I believe it can be proven in a straightforward way (and  using only what we have uptil now), that every map $\phi$ that we obtain be transforming a solvable system into a system of the above form is surjective (of course we don't know at this point, if we transform a system in two different ways to a system of the above form, whether for their associated maps $$\phi:\mathbb{R}^s\rightarrow L\subseteq \mathbb{R}^n \ \text{and} \ \phi:\mathbb{R}^{s'}\rightarrow L\subseteq \mathbb{R}^n$$we have $s=s'$ or $s\neq s'$. This is an essential point, I believe; with "later theory" it also follows that indeed $s=s'$, but we don't need that for surjectivity, nor that $r$ doesn't on the way we arrive from some system to the above system - in contrast to what the author said.)
(I assume I got so few comments because in the earlier version it wasn't clear, what I was asking. Now I hope it's more clear!)
 A: You are right that the author exaggerates the difficulty of proving that the map $\phi$ is surjective to the set $L$ of solutions, but your argument (in the comments) is not very clear, so I'll clarify it here.
The main point is that the system in echelon form is equivalent to the original system, so that the two have the same solution set $L$; this is implicit in the kind of transformations allowed to reach echelon form (no restrictions can be arbitrarily added or forgotten). So we just need to consider the echelon form system by itself.
The essence of the definition of $\phi$ is that it uses the equations of the system to deduce the values of all variables from the values of the free variables among them. Therefore $\phi(x_1,\ldots,x_s)\in L$ for all $(x_1,\ldots,x_s)\in\Bbb R^s$: one always gets a solution, whatever values one takes for the free variables. But we also know that $\phi$ did not change the values of the free variables themselves, so that if $i_1,\ldots,i_s\in\{1,\ldots,n\}$ are the positions of the free variables and $\pi:\Bbb R^n\to\Bbb R^s$ is the map extracting the values at those positions, so $\pi(x_1,\ldots,x_n)=(x_{i_1},\ldots,x_{i_s})$, then $\pi(\phi(x_1,\ldots,x_s))=(x_1,\ldots,x_s)$ (if we solve all variables from the free ones, but then forget those variables that aren't free, we just get back the values we started with). And finally we know that there was no choice in finding the non-free variables, since each one was deduced using an equation of the system from either the values of the free variables alone (for the first one solved) or from  the values of the free variables and previously solved values of non-free variables. This means that two solutions that have the same values for the free variables must be identical (the first non-free variable where they differ would contradict the fact that they are both solutions); in formula, for $(x_1,\ldots,x_n),(y_1,\ldots,y_n)\in L$ the condition $\pi(x_1,\ldots,x_n)=\pi(y_1,\ldots,y_n)$ (agreement on the free variables) implies $(x_1,\ldots,x_n)=(y_1,\ldots,y_n)$ (agreement everywhere).
Now one easily argues surjectivity of $\phi$. Let $(x_1,\ldots,x_n)\in L$ be any solution, and put $(y_1,\ldots,y_n)=\phi(\pi(x_1,\ldots,x_n))=\phi(x_{i_1},\ldots,x_{i_s})$. Then on one hand $(y_1,\ldots,y_n)$ is in the image of $\phi$, and therefore in $L$ ($\phi$ only produces solutions), and on the other hand $\pi(y_1,\ldots,y_n)=\pi(\phi(\pi(x_1,\ldots,x_n)))=\pi(x_1,\ldots,x_n)$. But we just saw that this implies $(x_1,\ldots,x_n)=(y_1,\ldots,y_n)$, and in particular $(x_1,\ldots,x_n)$ is (also) in the image of $\phi$. Stated without formulas: given any solution, we can extract from it the values of the free variables and construct from them using $\phi$ values for all the non-free variables; but since this is the only way to extend those values to a solution, we must have gotten back to our original solution.
