From the statement, $\sqrt[50]{x}=12 \quad \Rightarrow \quad x=12^{50}$ and $\sqrt[50]{y}=18 \quad \Rightarrow \quad y=18^{50}.$

We need the remainder when $x+y=12^{50}+18^{50}$ is divided by $25$.

So compute modulo $25$: $12^{50}+18^{50} \pmod{25}.$

Reduce the bases modulo $25$: $18 \equiv -7 \pmod{25}.$ Also, notice that powers modulo $25$ can be simplified using Euler's theorem: $\phi(25)=20.$ Since $\gcd(12,25)=1$ and $\gcd(18,25)=1$, $12^{20} \equiv 1 \pmod{25}, \qquad 18^{20} \equiv 1 \pmod{25}.$

Now, $50=20\cdot 2+10,$ so $12^{50}=12^{40}\cdot 12^{10} \equiv 1\cdot 12^{10}=12^{10} \pmod{25},$ $18^{50}=18^{40}\cdot 18^{10} \equiv 1\cdot 18^{10}=18^{10} \pmod{25}.$

Compute $12^{10} \pmod{25}$: $12^2=144 \equiv 19 \pmod{25}.$ Then $12^4 \equiv 19^2=361 \equiv 11 \pmod{25},$ $12^8 \equiv 11^2=121 \equiv 21 \pmod{25}.$ Therefore, $12^{10}=12^8\cdot 12^2 \equiv 21\cdot 19=399 \equiv 24 \pmod{25}.$

Compute $18^{10} \pmod{25}$: $18^2=324 \equiv 24 \pmod{25}.$ Then $18^4 \equiv 24^2=576 \equiv 1 \pmod{25}.$ Hence, $18^{10}=18^8\cdot 18^2=(18^4)^2\cdot 18^2 \equiv 1^2\cdot 24=24 \pmod{25}.$

Add the remainders: $12^{50}+18^{50} \equiv 24+24=48 \equiv 23 \pmod{25}.$

Therefore, the remainder on dividing $(x+y)$ by $25$ is $\boxed{23}.$