Intuitive reason for why $\int_1^{ab} \frac{1}{x} \, dx = \int_1^a \frac{1}{x} \, dx + \int_1^{b} \frac{1}{x} \, dx$?

Here is some hand-wavey intuition, which can be formalized with, say, substitution, or looking at partitions and upper or lower sums.

Subtract $\int_1^a\frac1tdt$ from both sides to get $$ \int_a^{ab}\frac1tdt = \int_1^b\frac1tdt $$ Why should this be true? If you move $a$ times further out along the $x$-axis, from $\int_1^b$ to $\int_a^{x}$, the graph is $a$ times closer to the $x$-axis, so you need to integrate over an $a$ times longer interval to get the same result. This would make $x = ab$.

Split $$\int_1^{ab}\frac1xdx=\int_1^a\frac1xdx+\int_a^{ab}\frac1xdx$$

Call $ay=x$, then $a\,dy=dx$ and, of course, $\displaystyle\frac1x=\frac{1}{ay}$.

Notice that $x=a\Longrightarrow y=1$ and $x=ab\Longrightarrow y=b$

Hence $$\int_a^{ab}\frac1xdx=\int_1^b\frac{1}{ay}\, a\,dy=\int_1^b\frac1ydy$$

The geometrical intuition you are looking for comes from this geometrical property of the integral above. Indeed, recall $x=y$ and you easily get

$$\int_1^{ab}\frac1xdx=\int_1^a\frac1xdx+\int_1^{b}\frac1xdx$$