How would multiple uses of a vigenère cipher be decrypted on paper (classic)?
My impression is that it would at most take a longer ciphertext to be decrypted. Imagine first the scenario where the different keys have the same length:

$K_1=hello$

$K_2=world$

In that scenario, the keys add up and the result of applying $E_{K_2}(E_{K_1}(m))$ can be expressed as one single operation: $E_{K*}(m)$, where $K*=dscwr$. Therefore, not a lot is two encryptions are nested with keys of the same length.

If you use keys of different lengths, you'll meet the same combination of letters at some point. Let's say you have a key of length 3 and a key of length five:

$K_3=abc$
$K_5=rolex$

After the position 15 ($5 \cdot 3$), you'll encrypt again a value with the same combination as in the position 1. Therefore, it is as if you had two 15-character keys:

$K_{15}'=abcabcabcabcabc$

$K_{15}*=rolexrolexrolex$

So, for this case in particular, you win nothing. 

Now, for the most general case, keep in mind that no paper cipher is a serious rival against the existent computational power. However, if by some clever narrative you manage to limit the powers of the adversary to just paper based attacks, there are paper algorithms much stronger than Vigenère. You can start by taking a look at Solitaire: https://www.schneier.com/cryptography/solitaire/