Vectors in $\mathbb{R}^2$ or $\mathbb{R}^3$ can be regarded (intuitively) as directed line segments. The word "directed" implies that direction/sense is important. The directed line segment $\vec{PQ}$ is not the same as the directed line segment $\vec{QP}$.
The vectors
$\mathbf{x} - \mathbf{y}$ and $\mathbf{y} - \mathbf{x}$ are not the same since they don't have the same direction. They have the same length, and they are parallel (some people would say "anti-parallel"), but they do not have the same sense.
Regarding the justification: if we allowed rotations, then any two vectors with the same length would be considered equal, so vectors would not be useful for describing directions. Even if we just allowed rotation by 180 degrees, every vector would be equal to its negative, and this implies that every vector is zero. So, requiring equality of length and direction (including sense) is the only option that makes sense.
I don't know of a name for the geometry arising from studying properties that are invariant under translation only.
The definition of linear independence says you can't make 0 out of a linear combination. It says nothing about not being able to make any other vector out of linear combinations.
(1,0) and (0,1) are independent since you cannot write (0,0) = c(1,0) + d(0,1) without c=d=0. But you can write every other vector as a nontrivial linear combination of these. (2,3) = 2(1,0)+3(0,1) for example. Spend some time making sense of the definitions with some concrete examples like this one and it will make sense eventually.
If you call your orthogonal set $\{v_1, v_2, \dots, v_n\}$, you can trivially write any vector in your set as a linear combination (take all coefficients $0$ except the coefficient of $v_k$ which is $1$).
$v_k = 0\cdot v_1+0\cdot v_2+\dots+0\cdot v_{k-1}+1\cdot v_k+0\cdot v_{k+1}+\dots+0\cdot v_n$
This is true of any set, whether it is orthogonal or not.
Moreover, any vector in the span of $\{v_1, v_2, \dots, v_n\}$ can be written as a linear combination of these vectors. This is again true of any set, whether orthogonal or not.
Best Answer
Hint: it suffices to prove that the diagonals of a rhombus are perpendicular. This follows from a basic fact from high school geometry, namely that the vertex of an isoceles triangle lies on the perpendicular bisector of the base.