Is this picture of a differential $1$-form correct? Consider the following picture of $x dy$:

(this picture appears in this article).
I believe these lines should be vertical:
At each point in $\mathbb R^2$  a basis for the tangent space is given by $dx, dy$. If $x$ and $y$ are the coordinates (local but in this case they are also global) then a point in $\mathbb R^2$ is given by $(x,y)$.
To get a basis for the tangent at a point one applies the partial derivatives and then evaluates at the point giving:
$$ dx = {\partial \over \partial x} (x,y) = (1,0)$$
and 
$$ dy = {\partial \over \partial y} (x,y) = (0,1)$$
Then 
$$ x dy = (0,x)$$
Which seems to me should be a vertical line. 

What am I doing wrong? Please could someone help me understand this?

 A: You are confusing an important point: the coordinate differentials $dx$ and $dy$ are (at each point) a basis of the cotangent space. What you are doing is identifying both spaces via the canonical inner product of $\mathbb R^2$, and saying $dy \mathrel{\hat=} \partial_y$. But then you miss an important point, namely that there is an important difference between forms and vectors. Their geometrical interpretations are straight lines (forms) and directed lines/arrows (vectors). Hence in your picture, the lines doesn't have arrows at the end! 
To make this picture clear, let us leave differential geometry behind, because this is not the essential ingredient here and only complicates things. Let us consider some vector space $V$ and its dual space $V^*$. Vectors $v\in V$ have a straight geometric interpretation: They are elements of the vector space and can be identified with directed line segments going from the origin 0 to $v$. But how can we interpret forms $f\in V^*$? These are linear maps $f:\,V\rightarrow\mathbb{R}$ and hence as long as $f\neq 0$, their kernels always have codimension 1. Hence it seems kind of natural to take this subspace as the geometric interpretation of $f$. Since its codimension is 1, it divides $V$ in half-spaces $f\leq 0$ and $f\geq 0$.
Back to the original setting: Here we have $V=\mathbb{R}^2$ and hence the mentioned kernels are 1-dimensional. Additionally, we now have to deal with differential forms and vector fields, which is essentially the same, except that we have to make every concept point-wise.
We want to compute the kernel of dy. The action of $dy$ on a vector field $X=X_x \partial_x + X_y \partial_y$ is given by $dy(X) = X_x dy(\partial_x) + X_y dy(\partial_y) = X_y$, so it picks out the $y$ component of $X$.$^1$ Then the solution of the equation $dy(X)=X_y=0$ are those vector fields which are proportional to $\partial_x$. This means that at each point, the kernel of $dy$ is just the $x$ axis of the tangent space. 
Finally, this is why we identify $dy$ with horizontal lines. Since $dy$ and $f(x,y)dy$ would result in the same line, we could say that we use a horizontal line, starting at $(x,y)$ which is $|f(x,y)|$ long and end up with your picture.
$^1$ Alternatively, this can be computed using the inner product via $dy(X) = \langle \partial_y, X \rangle = X_y$.
