In our course we discussed the following method to determine if given vectors are linear independent:
write each vector as a column of a homogenous matrix and evaluate if the matrix can be solved with a unique solution.
$$ \left[ \begin{array}{ccc|c} 1&0&0&0\\ 0&1&2&0 \end{array} \right] $$
But as far is I know I can only determine if a bunch of vectors is linear dependent/independent but I can not determine which of those are linear dependent. (note: Of course this example has a trivial, clear to see answer, but it could be not that obvious)
However if I use a different method by just writing each vector as a row:
$$ \left[ \begin{array}{ccc|c} 1&0\\ 0&1\\ 0&2 \end{array} \right] $$
I can easily try to reduce rows and can determine which vectors are linear independent. Is that true or am I missing some huge information so far?
Actually, you’ve gotten it backwards. Elementary row operations preserve linear dependency relationships among the columns of a matrix, not its rows. The fact that relationships among rows aren’t preserved should be pretty obvious from the echelon form: all of the dependent (zero) rows end up at the bottom regardless of which of the original rows are dependent on others. Consider, for example, the matrix $\begin{bmatrix}1&0\\1&0\\0&1\end{bmatrix}$.
So, assembling the vectors into a matrix as columns will let you pick out a linearly-independent subset of them that span the original set: the pivot columns of the reduced matrix tell you which those are. This won’t give you all possible independent subsets, though you can rearrange the vectors and re-reduce if you really want to do that.
On the other hand, assembling them into rows and reducing that matrix will give you a “nice” basis for their span, but in general it will consist of new vectors that weren’t in the original set.