What is the connection between a row vector and covariant vector (or column and contravariant)? This youtube video by Eugene Khutoryansky makes the distinction clear between the covariant coordinates of a vector (dot product of the vector with each of the basis vectors or vector projection), and the contravariant coordinates (parallelogram law):

Is there a way you could explain to a lay person that this somehow underpins the following fact:

Covariant vectors are representable as row vectors. Contravariant vectors are representable as column vectors.

?

I would like to know, for example, if the idea carries beyond being able to calculate the length of a vector in non-Cartesian coordinates as the dot product of its covariant and contr$avariant expressions: 
$$ \lVert V\rVert ^2=\begin{bmatrix} V_X & V_Y & V_Z\end{bmatrix}\cdot \begin{bmatrix} V^X \\ V^Y\\ V^Z\end{bmatrix}.$$
In a curvilinear system presumably the contravariant basis vectors would be tangential, whereas the covariant basis vectors would be orthogonal to the coordinates:


Apropos of the first comment, and if it can be confirmed (as a bonus), covariant vectors are covectors or dual vectors, while contravariant vectors are just vectors.
 A: You can represent "contravariant vectors" as rows and "covariant vectors" as columns all right if you want.
It's just a convention. The dual space of the space of column vectors can be naturally identified with the space of row vectors, because matrix multiplication can then correspond to the "pairing" between a "covariant vector" and a "contravariant vector".
Remember that "covariant vectors" are defined as scalar-valued linear maps on the space of "contravariant vectors", so if $\omega$ is a covariant vector and $v$ is a contravariant vector, then $\omega(v)$ is a real number that depends linearly on both $v$ and $\omega$. If you make $v$ correspond to a column vector, and make $\omega$ correspond to a row vector then $$ \omega(v)=\omega v=(\omega_1,...,\omega_n)\left(\begin{matrix}v^1 \\ \vdots \\ v^n\end{matrix}\right)=\omega_1v^1+...+\omega_nv^n. $$
If $\omega$ was the column instead, then the above matrix multiplication would look as $\omega(v)=v\omega$, which would not look as aesthetically pleasing, as we are used to displaying the argument of a function to the right of the function, and in this case $v$ is the argument.
