Determine the basis of the kernel and the image (matrix). But that's not the same? 
Let $f: \mathbb{R}^3 \rightarrow \mathbb{R}^2, f(x) = Ax$, where $A=
\begin{pmatrix} 1 & 2 & 3\\  4 & 5 & 6 \end{pmatrix}$
Determine the basis of the kernel $Ker(f)$ and the basis of the image
  $Im(f)$.

But I think it's enough if I just do it for one of these because it's actually the same..?
I transpose matrix $A$: $\begin{pmatrix}
1 & 4\\ 
2 & 5\\ 
3 & 6
\end{pmatrix}$ then I form it with Gauss (don't check for inaccuracies, it's correct) and get $\begin{pmatrix}
6 & 24\\ 
0 & 12\\ 
0 & 0
\end{pmatrix}$.
I transpose it back: $\begin{pmatrix}
6  & 0  & 0\\ 
24 & 12 & 0
\end{pmatrix}$
Thus $Im(f)= span\left(\left\{\begin{pmatrix}
6\\ 
24
\end{pmatrix}, \begin{pmatrix}
0\\ 
12
\end{pmatrix}\right\}\right)$, so Basis $B= \left\{\begin{pmatrix}
6\\ 
24
\end{pmatrix}, \begin{pmatrix}
0\\ 
12
\end{pmatrix}\right\}$
and that's also a basis of the kernel, isn't it?
 A: Gaussian elimination on $A$ gives
$$
\begin{bmatrix}
1 & 2 & 3 \\
4 & 5 & 6
\end{bmatrix}
\to
\begin{bmatrix}
1 & 2 & 3 \\
0 & -3 & -6
\end{bmatrix}
$$
which allows us to conclude that a basis for the image of $f$ is
$$
\left\{
\begin{bmatrix}1\\4\end{bmatrix},
\begin{bmatrix}2\\5\end{bmatrix}
\right\}
$$
However, since the matrix has rank $2$, any basis of $\mathbb{R}^2$ would do.
Backwards elimination goes on as
$$
\to
\begin{bmatrix}
1 & 2 & 3 \\
0 & 1 & 2
\end{bmatrix}
\to
\begin{bmatrix}
1 & 0 & -1 \\
0 & 1 & 2
\end{bmatrix}
$$
which allows us to say that a basis of the kernel is
$$
\left\{
\begin{bmatrix}1\\-2\\1\end{bmatrix}
\right\}
$$
Note that the image and the kernel live in different vector spaces, so they can't have the same basis.
With your method you're essentially doing column elimination, which certainly gives you a basis for the image. However, this cannot be used for going on and getting a basis for the kernel.
If instead you proceed with elimination on the matrix, the columns in the original matrix corresponding to the pivot column in the row echelon form give a basis for the image.
When you arrive to the reduced row echelon form, it's also easy to read off from it a basis for the kernel.
How do you do it? Interpret the RREF as the matrix of a homogeneous system; in our case it is
\begin{cases}
x_1-x_3=0\\
x_2+2x_3=0
\end{cases}
so we can give the value $x_3=1$ and compute the values for $x_1$ and $x_2$.
If the RREF is, for instance,
\begin{bmatrix}
1 & 2 & 0 & -1 \\
0 & 0 & 1 & 3 \\
0 & 0 & 0 & 0
\end{bmatrix}
the linear system is
$$
\begin{cases}
x_1+2x_2-x_4=0\\
x_3+3x_4=0
\end{cases}
$$
The free variables are $x_2$ and $x_4$, so we consider the vectors
$$
\begin{bmatrix}? \\ 1 \\ ? \\ 0\end{bmatrix}
\qquad
\begin{bmatrix}? \\ 0 \\ ? \\ 1\end{bmatrix}
$$
where $?$ denotes the value to be computed from the system. These two vectors are certainly linearly independent (why?). Thus we get
$$
\begin{bmatrix}-2 \\ 1 \\ 0 \\ 0\end{bmatrix}
\qquad
\begin{bmatrix}1 \\ 0 \\ -3 \\ 1\end{bmatrix}
$$
as the vectors forming a basis for the kernel.
