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 euklidean scalar product that consists of products of two 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

if you can get ahold of that.

Short answerts to the questions would then be:

1. Strictly speaking no, bivectors and two-forms live in different algebraic objects, but there is a canonical isomorphism of vector spaces.
2. In three dimensions there is a canonical isomorphism between the two, e.g. e_1 times 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 times e_2 means the product in the Clifford algebra build from that (as mentioned above).
3. If you write bivectors and vectors explicitly as elements of the Clifford algebra I mention above, the difference is manifest.
4. In 4-dim there is no canonical isomorphism, this works in 3-dim only (see itime 2).
5. A bivector can be visualized as a survace with a "direction" and a "size", e.g. in three dimensions a part of the x-y-plane plus "clockwise" or "counterclockwise". The vector would then be parallel to the z-axis, it's 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 times e_2 you get your vector by the familiar cross product of e_1 and e_2.
6. I do not know what "non-simple" means in this context, but maybe you think of elements like e_1 times e_2 + e_1 times e_3. That would be a bivector that consists of two elementary bivectors.

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:

1. Strictly speaking, no, bivectors and two-forms live in different algebraic objects, but there is a canonical isomorphism of vector spaces.

2. 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).

3. If you write bivectors and vectors explicitly as elements of the Clifford algebra I mention above, the difference is manifest.

4. In 4-dim there is no canonical isomorphism; this works in 3-dim only (see item 2).

5. 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 $e_2$.

6. 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.