How would I get an intuative idea of what this transformation does to the unit square? $$T(x,y)= (ye^x,2x)$$
I seem to think that the unit square is mapped to something that looks like

Would this be correct? 
The way I thought about this would be to fix $y$ then "trace" $x$. Does anyone have any other ways of doing this?
 A: Let us consider that the graphics you have provided is in the $(X,Y)$ image plane (uppercase X,Y versus lowercase $x,y$ for the initial space $[0,1] \times [0,1]$).
The relationships describing the transformation 
$$\tag{1} \cases{X=ye^x & (a) \\ Y=2x & (b)}$$
can be interpreted thus: we look for those $(X,Y)$ having a preimage $(x,y)$, i.e., being such that $\exists x,y \in [0,1]$ such that :
$$\tag{1} \cases{X=ye^x & (a) \\ Y=2x & (b)}$$
Let us (partially) inverse these relationships.
(1) is equivalent by taking the logarithm of relationship (1a):
$$\tag{2}\cases{\ln(X)=\ln(y)+\frac{Y}{2} & (a) \\ x=\frac{Y}{2} & (b)}$$
But condition $y \in [0,1]$ is equivalent to condition $\ln(y) \leq 0$.
Thus equation (2a) is equivalent to say that $\ln(X)<\frac{Y}{2}$.
Otherwise said, it means that all the region described by inequality: 
$$Y>2 \ln(X)$$
is convenient. 
But we must take into account the constraint brought by (2b): $0 \leq Y \leq 2$.
We have thus proved that the region you have found is the good locus (see picture below with the images of $2000$ random points in $[0,1] \times [0,1]$.

Remark: A completly different approach would have been to consider that the different constituents of the boundary of the image are the images of the four sides of the initial square.
For example, the image of the vertical side along the $y$ axis, that can be parameterized as the set of points $(x,y)=(0,t), \ \ t \in [0,1]$ is the set of points $(te^0,2 \times 0)=(t,0)$, i.e, is the horizontal side. Whereas the upper horizontal side which is the set of points $(x,y)=(t,1), \ \ t \in [0,1]$ is mapped onto the set of points $(1 \times e^t,2t)$ which is a parametric representation of the arc of logarithm curve.
A drawback of this method is that the region enclosed by the frontier curve is not necessarily mapped onto the region delimited by the image of the frontier. This has not happened here, but for the sake of rigor, it is better to work on inequalities and equivalent inequalities, as we have done in the first part.

