Double hashing constraints
In double hashing, we have (for a specific $k$) the series $h(k) + nh'(k) \mod m$
Since we have that $gcd(h'(k), m) = 1$ (they are relatively prime), then from Bézout's identity there exists two integers a,b such that $ah'(k) + bm = 1$. since $h'(k)$ is not 0, we can conclude that b is not 0 and thus also a is not 0.
We have $ah'(k) = 1 \mod m$, which means that a is the inverse of h'(k) in modulo m.
Now, let $0 \leq i \leq m$ be an integer. by choosing $n = ia$, we have $( h(k) + nh'(k) ) \mod m = ( h(k) + iah'(k) ) \mod m = ( h(k) + i ) \mod m$.
This concludes the proof (think why).
I hope i managed to help!