Others have given you very good answers for this specific question. Let me make a few points for general problems of this type:
If you have a linear transformation $f\colon\mathbb{R}^n\to\mathbb{R}^m$, since every vector in $\mathbb{R}^n$ can be written as a linear combination of the standard basis vectors: $\begin{array}{rcl} \mathbf{e}_{1} &=& (1,0,0,\ldots,0)\\ \mathbf{e}_{2} &=& (0,1,0,\ldots,0)\\ &\cdots&\\ \mathbf{e}_n &=& (0,0,0,\ldots,1), \end{array}$ then the span of $f$ equals $\langle f(\mathbf{e}_1,\ldots,f(\mathbf{e}_2)\rangle$. In the example you have, this is essentially what you did, since $f(1,0) = (1,2)$ and $f(0,1)=(-1,1)$. Second, to find the span of a set of vectors, you can use Gaussian elimination. Put them as rows of a matrix and perform Gaussian elimination (or Gauss-Jordan elimination). The span of the original vectors equals the span of the rows of the matrix, and elementary row operations does not change the span of the rows. The rank will give you the dimension of the image, and if you only want to know if the image is all of $\mathbb{R}^m$, then you just need to see if the rank of the reduced row echelon form is $m$.
But it is more common express the function as an $m\times n$ matrix: in the $i$th column, you write down the image of $\mathbf{e}_i$; the image of $f$ is then the span of the columns of this matrix. In your example, we would have $\left(\begin{array}{rr} 1 & -1\\ 2 & 1 \end{array}\right)$ (the first column is the image of $(1,0)$; the second column is the image of $(0,1)$).
Since the dimension of the columnspace is the same as the dimension of the rowspace, you can still use elementary row operations to figure out the rank. However, elementary row operations may change the columnspace (not the dimension of the columnspace, but the columnspace itself yes). So you have to be careful there. Here, it is very easy: subtracting twice the first row from the second row, then dividing the second row by $3$, and adding it to the first row we get: $\left(\begin{array}{rr} 1 & -1\\ 2 & 1 \end{array}\right) \to \left(\begin{array}{rr} 1 & -1\\ 0 & 3 \end{array}\right)\to\left(\begin{array}{rr} 1 & -1\\ 0 & 1 \end{array}\right) \to \left(\begin{array}{rr} 1 & 0\\ 0 & 1 \end{array}\right).$ Since the reduced row echelon form has rank $2$, the columnspace of the original has dimension $2$, so you conclude that $\dim(\mathrm{Im}(f)) = 2$. Since the only subspace of $\mathbb{R}^2$ that has dimension $2$ is $\mathbb{R}^2$ itself, you are done.