How does Euler's Theorem show that $(m^e)^d ≡ m \bmod n$
I thought that Euler's theorem basically proved that, for $(m^e)^d \equiv m \pmod n$ to work, $ed \equiv 1 \pmod {\phi(n)}$ must be true


No; a direct application of Euler's theorem shows that if $ed \equiv 1 \pmod {\phi(n)}$ is true (and $\gcd(m,n)=1$, Euler's theorem needs that as well), then we always have $(m^e)^d \equiv m \pmod n$

However, it does not imply the converse (and in fact, the converse is not true).

A stronger statement would be (assuming $n = pq$ for distinct primes $p, q$) that if $ed \equiv 1 \pmod{ p - 1}$ and $ed \equiv 1 \pmod{ q - 1}$, then we have $(m^e)^d \equiv m \pmod n$ for all $m$.

And, in this case, the converse is true; if we have $(m^e)^d \equiv m \pmod n$ for all $m$, then we necessarily have $ed \equiv 1 \pmod{ p - 1}$ and $ed \equiv 1 \pmod{ q - 1}$.  In fact, if we have either $ed \not\equiv 1 \pmod{ p - 1}$ or $ed \not\equiv 1 \pmod{ q - 1}$, then we'll necessarily have $(m^e)^d \not\equiv m \pmod n$ for at least 1/3 of the possible $m$ values.

Another (perhaps more common) way of writing these two equivalences is to express it as $ed \equiv 1 \pmod{ \lambda(n) }$ for the function $\lambda(n) = \text{lcm}(p-1, q-1)$.

How does these stronger statements relate to your original relation?  Well, if we have $ed \equiv 1 \pmod{\phi(n)}$, then we necessarily have $ed \equiv 1 \pmod{\lambda(n)}$ (and hence RSA "works")