How is the ln k term in the linearised Arrhenius equation acceptable?
The reason is because of sloppy notation. This sort of logarithmic plot is widespread in chemistry, though, and isn't limited to the Arrhenius equation, so do not expect it to change very much in the near future. You will have to get used to it.

With the Arrhenius equation

$$k = A \exp\left(-\frac{E_\mathrm a}{RT}\right)$$

you cannot directly take logarithms on both sides, because of the inherent units in $k$ and $A$. First you would have to divide through by a constant which we could call $k^\circ$:

$$\frac{k}{k^\circ} = \frac{A}{k^\circ} \, \exp\left(-\frac{E_\mathrm a}{RT}\right)$$

Now we can write

$$\ln \left( \frac{k}{k^\circ} \right) = \ln \left( \frac{A}{k^\circ} \right) - \frac{E_\mathrm a}{RT}$$

and as long as $k$, $k^\circ$, and $A$ are expressed in the same units, the resulting arguments of the logarithms will not have units. Note that the actual value of $k^\circ$ does not matter: you will of course get a different y-intercept, but the final values of $A$ and $E_\mathrm a$ that you get out of the Arrhenius plot will not be affected.

The division by $k^\circ$ is often implicit. When people write $\ln k$, what they usually mean is $\ln(k/k^\circ)$ where $k^\circ$ has the numerical value of $1$ and the units of whatever $k$ is in. Once you have the y-intercept, you take the exponential of that and tack the units back on to get $A$.

Although it is not rigorous, from a practical standpoint, this generally does not lead to much confusion as long as the author is consistent with their usage of units. Another reason for omission is that the actual value of $k^\circ$ has no physical effect, as previously discussed, and an overly pedantic focus on writing $k/k^\circ$ and $A/k^\circ$ can distract one from the chemical principles behind these equations, especially for more complicated equations which have many more symbols.