In RSA, why is it important to choose e so that it is coprime to φ(n)? When choosing the public exponent e, it is stressed that $e$ must be coprime to $\phi(n)$, i.e. $\gcd(\phi(n), e) = 1$. 

I know that a common choice is to have $e = 3$ (which requires a good padding scheme) or $e=65537$, which is slower but safer.

I also know that for two primes $p,q$, we have $\phi(pq) = (p - 1) (q - 1)$

Now, let me give a (simple) example:

Say I choose $e = 3$, and two random primes $p = 5$ and $q = 13$.

I can now compute $\gcd(3, \phi(5 \cdot 13)) = 3$.

This reveals that $3$ and $\phi(n)$ are not coprime. I assume this could also happen for large values of $p$ and $q$, and likewise for another $e$. I therefore assume that the RSA algorithm must check that $\gcd(e, \phi(pq)) = 1$. But let's assume it doesn't.

How does RSA become vulnerable if $\gcd(e, \phi(pq)) \neq 1$?
It doesn't become vulnerable; instead, it becomes impossible to decrypt uniquely.

Let us take the example you give: $N=65$ and $e=3$.

Then, if we encrypt the plaintext $2$, we get $2^3 \bmod 65 = 8$.

However, if we encrypt the plaintext $57$, we get $57^3 \bmod 65 = 8$

Hence, if we get the ciphertext $8$, we have no way of determining whether that corresponds to the plaintext $2$ or $57$ (or $32$, for that matter); all three plaintexts would convert into that one ciphertext value.

Making sure $e$ and $\phi(N)$ are relatively prime ensures this doesn't happen.

BTW: when you generate an RSA key, common practice nowadays is to select $e$ first, and then when you select the primes $p$, $q$, you make sure that $p-1, q-1$ are relatively prime to $e$; this is equivalent to making sure that $e$ and $\phi(N)$ are relatively prime.