Let $$D=\begin{pmatrix}1&2\\1&0\\1&0\end{pmatrix}.$$
I found the SVD to be $$U=\begin{pmatrix}0.9129&-0.0000&-0.4082\\0.1826&0.8944&0.4082\\0.3651&-0.4472&0.8165\end{pmatrix},$$ $$S=\begin{pmatrix}2.4495&0\\0&1.0000\\0&0\end{pmatrix},$$ $$V=\begin{pmatrix}0.4472&0.8944\\0.8944&-0.4472\end{pmatrix}.$$
I just do not know how to use this information to create a sketch of the range.
The range of the matrix S are the x and y axis in $\mathbb{R}^3$. U and V are rotation matrices. Any input to D is a vector in $\mathbb{R}^2$. Let $\{e_1,e_2\}$ be the standard basis for $\mathbb{R}^2$. Now consider a separate matrix $\beta=V^{-1}[\begin{array}{cc} e_1 & e_2 \end{array}]$. What is the image of this matrix?
$V\beta$ is just the identity matrix. So $S V \beta$ is just $S$. $U$ is a rotation matrix of S. So these axis get rotated into $\mathbb{R}^3$. So the image of $D$ equal to the image of $U S V $ is the span of the columns of $U S V \beta$. This is a plane in $\mathbb{R}^3$.