Theorem: a polynomial $f$ over $K$ has a repeated zero in its splitting field if and only if $f$ shares a common factor with $f'$ in $K[t]$.
The author starts by proving that if $f$ has a repeated zero, then $f$ and $f'$ share a common factor of degree $≥1$ in $K[t]$, in the following way:
$$f=(t-a)^2g$$ $$f'=(t-a)^2g'+2(t-a)g$$ $$=(t-a)[(t-a)g'+2g]$$
He follows to claim that "since $f$ and $f'$ share a common factor in $\Sigma[t]$, they must share a common factor in $K[t]$", where $\Sigma$ is the splitting field of $f$ over $K$.
I understand how, from the fact that $f$ has a repeated zero in its splitting field, $f$ and $f'$ must share a factor in $\Sigma[t]$, but how does the author infer from this that they must then share a factor in $K[t]$?
The second part of the proof, in which the author proves that if $f$ has no repeated zeroes then it doesn't share a common factor with $f'$, is understandable since it doesn't depend on $K[t]$.
I would really appreciate any help/thoughts.
The reference to the splitting field could be misleading. Suppose there is an extension field $F$ where $f$ has a multiple root $a$.
The argument in your question proves that $f$ and $f'$ have a common root. In particular, their greatest common divisor $p$ in $F[t]$ has degree $>0$.
Now the key fact is that, since both polynomials belong to $K[t]$, their gcd is computed with the Euclidean algorithm that takes place in $K[t]$. If you think to compute it in $F[t]$, you would do exactly the same operations; hence the gcd is the same in $F[t]$ as in $K[t]$.
This proves that $p\in K[t]$ and, since it has positive degree, it is the required nonconstant common factor.