In one of my problems it says the following: Let $K$ be an infinite field. Consider the linear action of $GL_2$ on $K[x,y]$ induced by the natural representation of $GL_2$ on $K^2$.
I don't know what the first part means. I know what $GL_2$ looks like, I also know what $K[x,y]$ looks like. I think "the natural representation of $GL_2$ on $K^2$" means that we look at $GL_2$ as $2x2$ matrices, with entries in $K$, and let that act in the natural way (i.e. left multiplication) on the vector space $K^2$. But how does that induce an action on $K[x,y]$. For example I don't know how the image of $x$ would look like under that action, or in general how that induced action looks like.
You are correct about what the natural representation of $\mathrm{GL}_2$ on $K^2$ is. You can think of $K[x,y]$ as the space of polynomial functions on $K^2$, where $x$ is the function returning the first coordinate and $y$ returns the second, so
$$x(a,b)=a,\:y(a,b)=b$$
Now you can define $x\cdot g(a,b)=x(g\cdot(a,b))$ for each $g\in\mathrm{GL}_2$, and similarly for $y$, to give an action of $\mathrm{GL}_2$ on $K[x,y]$. Explicitly, we have
$$\begin{pmatrix}g_{11}&g_{12}\\g_{21}&g_{22}\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix}=\begin{pmatrix}g_{11}a+g_{12}b\\g_{21}a+g_{22}b\end{pmatrix}$$ so for any $(a,b)\in K^2$, we have
$$x\cdot g(a,b)=g_{11}a+g_{12}b=g_{11}x(a,b)+g_{12}y(a,b)$$
Thus $x\cdot g=g_{11}x+g_{12}y$. Similarly $y\cdot g=g_{21}x+g_{22}y$.
This turns the left action on $K^2$ into a right action on the space of functions $K[x,y]$ - if you want to stick to left actions, you will need to define $g\cdot x(a,b)=x(g^{-1}\cdot(a,b))$ instead. I hope the right choice is clear from your context.