Please note that I am very new to Linear Algebra and am having some confusion regarding orthogonal vectors.
Let $A = \begin{bmatrix}1 & -1\\ 2 & 2\end{bmatrix}$
My current understanding is that $A$ represents a transformation to another space (call it destination space) where the ortho-normal basis vectors would be in the direction of the above two column vectors $\begin{bmatrix}1\\ 2 \end{bmatrix}$ and $\begin{bmatrix}-1\\ 2 \end{bmatrix}$.
So for someone doing linear algebra in destination space, their equivalents to our orthonormal basis $b_1 =\begin{bmatrix}1\\ 0 \end{bmatrix}$ and $b_2 =\begin{bmatrix}0\\ 1 \end{bmatrix}$ would be $ \begin{bmatrix}1\\ 2 \end{bmatrix}$ and $\begin{bmatrix}-1\\ 2 \end{bmatrix}$ since $Ab_1 = \begin{bmatrix}1 \\ 2 \end{bmatrix}$ and $Ab_2 = \begin{bmatrix}-1 \\ 2 \end{bmatrix}$.
Now, if we take the dot-product of these two vectors when viewed from destination space, the result would be 0. Does this mean that these two vectors, $\begin{bmatrix}1\\ 2 \end{bmatrix}$ and $\begin{bmatrix}-1\\ 2 \end{bmatrix}$ would be considered to be orthogonal in destination space?
So the angle between two orthogonal vectors would be different in spaces with different basis vectors?
The answer depends to a large extent on your point of view as to what’s going on with the two vectors spaces and the transformation. However, what all of the answers have in common is that orthogonality of vectors is defined with respect to some specific inner product $\langle\cdot,\cdot\rangle$: $\mathbf v$ and $\mathbf w$ are orthogonal iff $\langle\mathbf v,\mathbf w\rangle = 0$. So really, once we’re asking about orthogonality, we’re not working with bare vector spaces but with inner product spaces—vectors spaces equipped with an inner product. More broadly, the angle $\theta$ between a pair of vectors is related to a specific inner product via that formula $\langle\mathbf v,\mathbf w\rangle = \langle\mathbf v,\mathbf v\rangle \langle\mathbf w,\mathbf w\rangle \cos\theta$.
If you have a linear transformation $T:V\to W$, where both $V$ and $W$ are inner product spaces, it almost self-evident that there’s no particular reason to expect that the linear transformation also preserves inner products of vectors. That is, in general $\langle T\mathbf v_1,T\mathbf v_2\rangle_W\ne\langle\mathbf v_1,\mathbf v_2\rangle_V$. This is what gimusi is referring to in his answer when he says that linear transformations don’t in general preserve angles.
It’s important to distinguish between an inner product, which is a function on pairs of vectors, and the dot product, which is a function of the coordinates of a pair of vectors. The latter only operates on tuples of scalars, whereas it makes perfect sense to define inner products on spaces in which the vectors are functions. Now, when you’re working with $\mathbb R^n$ equipped with the Euclidean scalar product $(\cdot,\cdot)$, and you’re working in the standard basis $\mathcal E$, then these two coincide: $(\mathbf v,\mathbf w) = [\mathbf v]_{\mathcal E}\cdot[\mathbf w]_{\mathcal E} = [\mathbf v]_{\mathcal E}^T[\mathbf w]_{\mathcal E}$. This doesn’t hold for any basis of $\mathbb R^n$, though. For example, in $\mathbb R^2$, the standard basis vectors are orthogonal $(\mathbf e_1,\mathbf e_2)= [\mathbf e_1]_{\mathcal E}\cdot[\mathbf e_2]_{\mathcal E} = (1,0)^T\cdot(0,1)^T = 0$. On the other hand, in the ordered basis $\mathcal B = \left((1/2,-1/2)^T,(1/4,1/4)^T\right)$, $[\mathbf e_1]_{\mathcal B}\cdot[\mathbf e_2]_{\mathcal B} = (1,2)^T\cdot(-1,2)^T = 3\ne 0$. This is algebraically equivalent to your example, but the point of view is a bit different. Instead of the linear transformation mapping one Euclidean space to another and working in the standard basis of both spaces, we have a transformation that converts coordinates in one basis to another within the same Euclidean space. The vectors themselves don’t change, so the angle between them shouldn’t change, either. What needs to change is the coordinate-based formula for their inner product.
The necessary adjustment to the Euclidean inner product formula is easily computed from the change of basis matrix $M$: $$[\mathbf v]_{\mathcal E}^T[\mathbf w]_{\mathcal E} = (M^{-1}[\mathbf v]_{\mathcal B})^T(M^{-1}[\mathbf w]_{\mathcal B}) = [\mathbf v]_{\mathcal B}^T(M^{-T}M^{-1})[\mathbf w]_{\mathcal B}.$$ It turns out that every scalar product $\langle\mathbf v,\mathbf w\rangle$ on a finite-dimensional vector space can be expressed in $\mathcal B$-coordinates as $[\mathbf v]_{\mathcal B}^TA[\mathbf w]_{\mathcal B}$, where $A$ is some symmetric matrix and when you change bases via the matrix $M$, the matrix $A$ in this formula becomes $M^{-T}AM^{-1}$.