Are there groups $G$, whose order is uncountable, such that the order of $G/[G,G]$ is countable?

Yes, in the group $PSL(n, \mathbb{C})$ every element is a commutator, so the quotient has cardinality $1.$ A weirder example is the permutation group $S_\Omega$ of an infinite set $\Omega.$

As I said in my comment to Igor Rivin's answer, if $\kappa$ is an infinite cardinal then the group $A_{\kappa}$ of even permutations with finite support is simple. Therefore, the abelianisation of $A_{\kappa}$ is trivial.

In fact, as $A_{\kappa}$ has cardinality $\kappa$ this yields an example for every possible cardinality.

To see that $A_{\kappa}$ is simple, apply the following lemma to the fact that $A_n$ is simple for $n\geq 5$. (See page 73 of D.J.S. Robinson's book A course in the theory of groups.)

Lemma: If $G$ is the union of a chain of simple groups, so $G=\cup H_i$ where $H_0\leq H_2\leq H_3\leq\ldots$ and each $H_i$ is simple, then $G$ is simple.

Proof: Suppose $1\neq N\unlhd G$. We shall prove that $N=G$. Because $G$ is the union of a chain of simple groups there exists some simple group $H_i$ with $N\cap H_i\neq 1$, and then $N\cap H_j\neq 1$ for all $j>i$. But then $N\cap H_j\lhd H_j$ so $H_j\leq N$ for all $j>i$. We thus conclude that $N=G$, as required.