Dan's right, the inclusion $K(Y)\subset K(X)$ would give a map $X\to Y$. However, I think you meant to write $K(X)\subset K(Y)$, which is easier to see is the correct formulation if you write it as $K(u,v)\subset K(t)$. Also, you meant to say $s$ as a rational function of $u$ and $v$, not of $X$ and $Y$.


You can get an equation (probably singular) for $X$ by computing the resultant of $(1+4t)^5u-t(4+t)^5$ and $(1+4t)v-t^5(4+t)$ with respect to the variable $t$. This gives an equation in $u$ and $v$. After dividing by a common factor, I get that $X$ is given by the following rather horrible affine equation in the $(u,v)$-plane:
$$\eqalign{
-u^6
 &+ (281474976710656 v^5
 - 24739011624960 v^4
 + 685517045760 v^3 \cr
 &- 6305218560 v^2
 + 11887110 v)  u^5
 + (-24739011624960 v^5\cr
 &+ 301150356111360 v^4
 - 292164004085760 v^3
 - 7564564531215 v^2\cr
 &- 6305218560 v)  u^4
 + (685517045760 v^5
 - 292164004085760 v^4 \cr
 &+ 114773655178260 v^3
 - 292164004085760 v^2
 + 685517045760 v)  u^3 \cr
 &+ (-6305218560 v^5
 - 7564564531215 v^4
 - 292164004085760 v^3 \cr
 &+ 301150356111360 v^2
 - 24739011624960 v)  u^2
 + (11887110 v^5 \cr
 &- 6305218560 v^4
 + 685517045760 v^3
 - 24739011624960 v^2 \cr
 &+ 281474976710656 v)  u
 - v^6 = 0.
  }
$$
Then $t\to(u(t),v(t))$ gives the map from $\mathbb{P}^1=Y$ to $X$. Someone who is more adept than I am using computer algebra systems can probably figure out the degree of this map, and write $k(u(t),v(t))$ as $k(s(u(t),v(t)))$ for an explicit $s(u,v)$.