Why does orthogonal basises makes it easier to solve SVP in lattices?
Is it only because it somehow makes it easier for algorithms such as LLL(-BKZ) to reduce the basis vectors?

You have this somewhat backwards --- the point of basis reduction is to find a basis that is &quot;shorter&quot; and &quot;more orthogonal&quot;, which you then use for useful applications.
Babai's nearest planes is a fairly well-known algorithm to solve an approximate form of CVP, that is defined relative to some fixed basis $B = [b_1,\dots,b_n]$ of a lattice.
If $x\in\mathbb{R}^n$, and:

$v$ is the output of Babai's nearest planes, and
$v_{opt}$ is the closest vector to $x$,

then one can prove that:
$$\lVert v-v_{opt}\rVert_2^2 \leq \frac{1}{4}\sum_{i = 1}^n \lVert b_i^*\rVert_2^2$$
where $\{b_1^*,\dots, b_n^*\}$ is the Gram-Schmidt orthogonalization of $B$.
Note that as the Gram-Schmidt orthogonalization applies solely elementary operations, one has that:
$$\det([b_1^*,\dots, b_n^*]) = \det B$$
e.g. the total product $\prod_{i=1}^n \lVert b_i^*\rVert_2 = \det B$ is kept invariant.
It should be easy to see that the error bound in Babai's nearest planes will be minimized if this &quot;$\det B$ amount of mass&quot; is &quot;spread out evenly&quot; through all of $\{b_1^*,\dots, b_n^*\}$, e.g. if the basis is short and nearly orthogonal.