The big questions behind are:
A bivector is an element of $\bigwedge^2 V$, so it is dual to a $2$-form on $V$. You can think of a bi-vector as a tiny piece of area.
If $V$ is three dimensional and comes with an inner product, then one can choose an isomorphism between $V$ and $\bigwedge^2 V$ which commutes with all the orthogonal maps for your inner product. In elementary math, this is the map which we call the cross product. This is not quite unique; you have to decide whether to use the left-hand-rule or the right-hand-rule to take cross products.
In my opinion, the best way to learn to distinguish between vector and bivectors is to get in the habit of not identifying $V$ and $V^*$. One way to do this is to work with an inner product given by an arbitrary symmetric matrix $g$ and keep the matrix $g$ in all your computations, rather than changing to an orthonormal basis.
A quicker way which I find useful is to think about whether the quantity in question has a natural direction, or has a sign ambiguity which comes from some arbitrary convention. For example, the normal vector to an orientated surface in $3$ space is going to be a bivector, because we need to decide whether the orientation circles the normal to the left or the right.
Writing down a bi-vector in $d$ dimensions takes $\binom{d}{2}$ coordinates. So, for $d = 4$, we need $6$ coordinates and we can't fit them into a single vector. I'm guessing that "simple" means a wedge of two vectors. So $e_1 \wedge e_2 + e_3 \wedge e_4$ is not simple. Once we get up into higher than $3$ dimensions, there is nothing that prevents this, so it can happen.
Metaquestion to our hosts: Is there another way to post a follow-up question than postion it as an answer to the first one? To Pedro: Ok, now I get what you mean by simple bivector. Your question is answered in John Baez' blog:
http://math.ucr.edu/home/baez/week120.html
(Just do a text search for "bivector").
I believe the answer to your question on simplicity is no, e.g. $(1,0,0,0)\wedge (0,0,1,0) + (0,1,0,0)\wedge (0,0,0,1)$ cannot be written in the form $f\wedge g$.
One possible interpretation of the question uses Clifford algebras: A bivector could be defined as an element of the Clifford algebra of the $n$-dimensional real vector space with the Euclidean scalar product that consists of products of two orthogonal elements.
Have a look at the Wikipedia entry and the book
Lounesto, Pertti (2001), Clifford algebras and spinors, Cambridge: Cambridge University Press, ISBN 978-0-521-00551-7 (MR)
if you can get ahold of that.
Short answers to the questions would then be:
Strictly speaking, no, bivectors and two-forms live in different algebraic objects, but there is a canonical isomorphism of vector spaces.
In three dimensions there is a canonical isomorphism between the two, e.g., $e_1 \wedge e_2$ (bivector) is taken to $e_3$ (vector). The $e_1$ etc. are the elements of the canonical basis of your vector space, $e_1 \wedge e_2$ means the product in the Clifford algebra built from that (as mentioned above).
If you write bivectors and vectors explicitly as elements of the Clifford algebra I mention above, the difference is manifest.
In 4-dim there is no canonical isomorphism; this works in 3-dim only (see item 2).
A bivector can be visualized as a surface with a "direction" and a "size", e.g., in three dimensions a part of the $xy$-plane plus "clockwise" or "counterclockwise". The vector would then be parallel to the $z$-axis, its length equal to the size of the bivector. It's pretty easy to draw, but hard to describe with words. But if you write the bivector as, e.g., $e_1 \wedge e_2$, you get your vector by the familiar cross product of $e_1$ and $e_2$.
I do not know what "non-simple" means in this context, but maybe you think of elements like $e_1 \wedge e_2 + e_1 \wedge e_3$. That would be a bivector that consists of two elementary bivectors.
Tim's answer was mostly good, but I have to point out one thing: the isomorphism between bivectors and vectors in dimension 3 is not canonical; it depends on choosing something like a basis (as Tim did) or a Riemannian metric.
Thank you for all your answers, now I can ask the true question: A 4-simplex (in R^4) determines a set of 10 surfaces (true ones i.e: 2-simplices) so it determines also a set of 10 bivectors, for a 4-simplex to be uniquely determined by a set of 10 bivectors (up to parallel translation and inversion through the origin) the later must satisfy some conditions, among these conditions we find: (2) Each bivector is simple, i.e. of the form b = f ∧ g. (Are there bivectors which are not simple? The simplicity condition is encoded in their own definition!!!)
Ok so condition (2) simply says that each simple bivector (by its simplicity) will define two vectors thus will define a 2D surface, thus we have constructed the 10 surfaces of the future 4-simplex by this requirement.
I'm not sure if I understand you, but I'll try :-)
You are handed a set of 10 bivectors that live in four (real) dimensions. You want to find out if there is a simplex and an orientation on each of it's faces, such that the surfaces of the simplex are represented by your bivectors.
A necessary condition is that each of the bivectors is simple. Why? Answer: In 4 dimensions it could actually happen that some of the bivectors are not simple. That cannot happen e.g. in 3 dimensions. A bivector that is not simple does not represent a surface, therefore the condition is necessary.
But this condition alone is not sufficient to ensure that a set of 10 bivectors represent faces of a simplex. There are more (see my link to the blog of John Baez).
