I'm not sure what you are trying to do, and I fear you are misreading the text you are quoting. Changing basis amounts to going from the x-basis to the k-basis, which you wisely chose to replicate/parallel on your $\mathbb{R}^2(\mathbb{R})$ vector space.
The normalized x-basis is then simulated by $|e_1\rangle = (1, 0)^T$ and $|e_2\rangle = (0, 1)^T$. The k-basis is simulated by $|d_1\rangle = (1, 1 )^T/\sqrt{2}$ and $|d_2\rangle = (-1, 1)/\sqrt{2}$. The quantum state $|f\rangle$ is simulated by $|\vec{v}\rangle = -(1, 1)^T=-\sqrt{2} |d_1\rangle $.
To change basis from x to k, as in your (1.10.34), you utilize the (orthogonal; in QM it is unitary) transformation matrix T,
$$
|d_i\rangle = |e_1\rangle \langle e_1 |d_i\rangle + |e_2\rangle \langle e_2 |d_i\rangle = \sum_j T_{ij} |e_j\rangle, \\ T_{ij}=\langle e_j|d_i\rangle, \\
|\vec d\rangle \langle \vec e|= {\mathbb T}= \begin{pmatrix} 1&1 \\ -1 & 1 \end{pmatrix}/\sqrt{2}, \tag{1}
$$
where ${\mathbb T}^{-1} = {\mathbb T}^{T}$ taking you back from the d to the e basis.
Importantly, the two bases are complete, so, as above, (and in your text, where their continuous analogs are inserted),
$$
{\mathbb I}= |e_1\rangle \langle e_1|+ |e_2\rangle \langle e_2| = |d_1\rangle \langle d_1| + |d_2\rangle \langle d_2|. \tag{2}
$$
You may now compute the "wavefunctions",
$$
\langle e_i|\vec v\rangle = -(1,1)^T \\ \langle d_i|\vec v\rangle = \sum_j \langle d_i|e_j\rangle \langle e_j|\vec v\rangle = - \sqrt{2} \delta_{i1}. \tag{3}
$$
I've no idea what the latter part of your questions purports to be doing. The bases' relation is valid for all representations. You already changed bases. This generalizes naturally to QM.
Best Answer
No, not always. Consider in $\Bbb{R}^n$ the three vectors: \begin{align*} e_1 &= (1, 0, 0) \\ e_2 &= (-1, 0, 1) \\ e_3 &= (-1, 0, -1). \end{align*} Then \begin{align*} e_1 \cdot e_2 &= -1 \\ e_1 \cdot e_3 &= -1 \\ e_2 \cdot e_3 &= 0. \end{align*} If there were unit vectors $e'_1, e'_2, e'_3$ with the desired properties, then $$e'_1 \cdot e'_2 = e'_1 \cdot e'_3 = -1,$$ which would imply that $e'_2 = e'_3 = -e'_1$. We would therefore be forced to have $e'_2 \cdot e'_3 = 1 \neq 0$.