How to express the Euclidean basis vector $(0, 1)$ under this new basis? Changing our basis. Suppose we are working in the $\mathbb{R}^2(\mathbb{R})$ vector space under the standard Euclidean basis and dot product.  Suppose we then transform the standard basis vectors $e_1 = (1,0) \mapsto (1/\sqrt{2}, 1/\sqrt{2}) = d_1$ and $e_2 = (0,1) \mapsto (-1/\sqrt{2}, 1/\sqrt{2})= d_2$.  Note that this is an orthogonal change in basis (in fact a $90$ degree rotation), so norms should be preserved.
How to re-express $e_2$ under the new $d_i$ basis? Now let us consider $e_2$ under the new $d_i$ basis.  It seems to me $e_2 = (1, 1)$ under this new basis, since
$$
1(d_1) + 1(d_2) = e_2
$$
Yielding a contradiction. But doesn't this then imply that
$$
||e_2|| \ne 1
$$
which is false? To see why, consider the expansion of $||e_2||$:
$$
||e_2|| = \sqrt{e_2 \cdot e_2} = \sqrt{(1,1) \cdot (1,1)} = \sqrt{1*1 + 1*1} = \sqrt{2} \ne 1
$$
So what I have done wrong? Note that under the standard basis, $||e_2|| = 1$ since
$$
||e_2|| = \sqrt{e_2 \cdot e_2} = \sqrt{(0,1) \cdot (0, 1)} = \sqrt{0*0+1*1} = 1
$$
...so this is clearly not right.

Your basis change is
$$d_1=\frac{1}{\sqrt 2}e_1+\frac{1}{\sqrt 2}e_2,$$
$$d_2=-\frac{1}{\sqrt 2}e_1+\frac{1}{\sqrt 2}e_2.$$
Solving for $e_2$ requires to do
\begin{eqnarray*}
d_1+d_2&=&\frac{1}{\sqrt 2}e_1+\frac{1}{\sqrt 2}e_2
+\left(-\frac{1}{\sqrt 2}e_1+\frac{1}{\sqrt 2}e_2\right),\\
&=&\frac{1}{\sqrt 2}e_1+\frac{1}{\sqrt 2}e_2
-\frac{1}{\sqrt 2}e_1+\frac{1}{\sqrt 2}e_2,\\
&=&\frac{2}{\sqrt 2}e_2.
\end{eqnarray*}
Then
$$e_2=\frac{\sqrt 2}{2}d_1+\frac{\sqrt 2}{2}d_2.$$
Also, here we can see $\|e_2\|=1$
