What does degrees of freedom mean Say I have a vector space V with dim V=n.
Explain the following to me in terms of degrees of freedom:
Any set of n linearly independent vectors from V spans V
Any set of n-1 linearly independent vectors from V doesn’t necessarily span a subspace of dimension n-1
Any set of n-2 linearly independent vectors from V doesn’t necessarily span a sub space of V of dimension n-2
E.g. In $\mathbb R^3$, take the sub space spanned by (1,0,0) and (0,1,0). (1,0,0) and (0,0,1) does not span this, even though the sub spaces are the same dimension. I would like someone to elucidate this in terms of the concept “degrees of freedom”
 A: At each point in $\Bbb{R}^3$, there are three degrees of freedom -- that is for each point of the space, we can setup up a (typically "small", because not all spaces are as nice and flat as Euclidean spaces) local coordinate system of displacements from that point which looks like a copy of $\Bbb{R}^3$.  (Note that the local coordinates of displacements do not have to run parallel to the coordinates in the space.  In fact, it is sometimes useful to have the local coordinates conform to some geometric object which we have placed in the big space and these need not be parallel to the big space's coordinates.)
A vector subspace of $\Bbb{R}^3$ is either the origin, a line through the origin, a plane through the origin, or all of $\Bbb{R}^3$ (which we won't discuss further since you seem to be interested in proper subspaces).  If we restrict to one of these (proper subspaces), we have fewer than three degrees of freedom.

*

*For the origin, a $0$-dimenaional subspace, we need $0$ local coordinates to represent possible displacements.  That is, since any displacement violates restriction to the subspace, no displacement is possible and no local coordinates of displacement are needed.

*For any point on a line through the origin, one only needs one local coordinate to represent displacements along the line, so there is only one degree of freedom.  Note that most lines through the origin are not parallel to the big space's coordinate axes, so the displacement coordinates are not necessarily the same as the big space's coordinates.

*For a point on a plane through the origin, one only needs two local coordinates to represent displacements in the plane, so there are only two degrees of freedom.  Again, the local coordinates need not be parallel to the big space's coordinates.

You talk about the subspace spanned by $(1,0,0)$ and $(0,1,0)$.  This is a plane through the origin.  In particular, it is the $xy$-plane, the plane of points in $\Bbb{R}^3$ with $z$-coordinates equal to zero.  Each point on this subspace has two degrees of freedom.  We are free to choose our local coordinates to point in any direction.  We can make the parallel to and the same scale as the $\Bbb{R}^3$ coordinates, but recall that displacements are measured from the point where you start, not from the origin, so the only point on the plane that can actually have its displacement coordinates exactly agree with the $\Bbb{R}^3$ coordinates is the origin.  For other points, their local coordinates must be displaced to put the origin at their location.
You also talk about the subspace spanned by $(1,0,0)$ and $(0,0,1)$.  This is the $xz$-plane, that is the plane of all points in $\Bbb{R}^3$ with $y$-coordinate equal to zero.  Similar comments about degrees of freedom and displacement coordinates apply on this plane as for the $xy$-plane.
The intersection of these two planes is the set of all points having $z$-coordinates equal to zero and $y$-coordinates equal to zero.  So the intersection of these two planes is the $x$-axis.  Every point in both planes has a degree of freedom parallel to this intersection, so every point of both spaces can have one of its local coordinate axes run parallel to the $x$-axis.  However, the second degree of freedom direction at any such point in the $xy$-plane is not a degree of freedom direction on the $xz$-plane, so there is no choice of local coordinates for points on one plane that corresponds to a possible displacement on the other plane.  Certainly all such points can have one degree of freedom direction direction along the $x$-axis, but their second directions cannot be chosen the same.
In particular, one can take the displacement vector $(1,0,0)$ to be a degree of freedom direction for every point of both planes.  However, on the $xy$-plane, the other displacement vector at each point must be chosen from the $xy$-plane minus the $x$-axis.  ("minus the $x$-axis" because we already have a local coordinate axis pointing in that direction, and we need our coordinates to be linearly independent.)  For the $xz$-plane, for each point, the second degree of freedom direction must be chosen from the $xz$-plane subspace minus the $x$-axis.
But this means that the local coordinate directions for a point on one plane are not parallel to vectors in  $\Bbb{R}^3$ that span the other plane.
It is easy enough for two subspaces to have a nonempty intersection.  (In fact the intersection of any pair of subspaces necessarily includes the origin.)  Consider two distinct random lines in the plane that intersect at the origin.  These are one-dimensional subspaces and if we examine the restrictions to either line, we find one degree of freedom for each point in each line.  But a vector that spans one line does not even lie in the other, so cannot span the other.  A similar thing happens with the two planes you call out: they intersect, but there are vectors in one that are not in the other and vice versa, so neither can span the other.
If vector (sub-)space $V$ contains a spanning set for vector (sub-)space $W$, then necessarily $V$ contains $W$, so their intersection is all of $W$.  If two (sub-)spaces have nonempty intersection, but the intersection does not contain all of either space, then a spanning set for one cannot span the other and vice versa.
A: For a set of vectors to span a subspace, the vectors need to be contained in that subspace.
Take your example: In $\mathbb{R}^3$ the subspace spanned by $(1, 0, 0)$ and $(0, 1, 0)$, say $W$ is not the same as the subspace spanned by $(1, 0, 0)$ and $(0, 0, 1)$.
For example the vector $(0, 0, 1)$ is not an element of $W$, because you can't write it as a linear combination of $a(1, 0, 0) + b(0, 1, 0)$.
The two spaces are isomorphic, though and have the same dimension.
This doesn't really relate to degrees of freedom in any way.
