Lower bound for sorting n arrays of size k each
For the first part, as the comments suggest - observe that there are $k!^n$ possible permutations for the inputs - indeed, every array has $k!$ permutations, and there are $n$ arrays, so you have $k!$ options for the first array, $k!$ for the second, and so on.

So the height of a decision tree is 
$$\log (k!^n)=n\log (k!)\in \theta(nk\log k)$$
So you need at least that many comparisons in order to determine a single leaf. 

For the second part - the argument you suggest is quite close to a formal explanation.
The lower bound of $\Omega(k\log k)$ is for comparison-based algorithms that compare elements in the array. When you have several arrays, then it is not justified to assume that sorting them is equivalent to sorting each one independently. 

A-priori, it may be easier to sort them by comparing elements between the arrays. But the lower bound on sorting does not utilize such a setting, so it is not directly relevant for this problem. 

In my opinion, this is as formal as you can get without formally defining what a comparison-based algorithm actually is. To define the latter, you need to formally model the computational model you use (which would be a sort of pointer-machine), and this is unlikely to result in any insights other than what you already mentioned.