$K[x]$ means the polynomial ring over $K$. You can prove $K[X]/(x^2)$ is isomorphic to the direct sum $K\dot{+}K\bar{x}$, where $\bar{x}^2=0$, by thinking about the homomorphism $$\phi:K[X]\to K\dot{+}K\bar{x}$$ sending $p(x)$ to $p(\bar{x})$, which is also just the remainder of $p(x)$ when divided by $x^2$ (except with $x$ replaced by $\bar{x}$). The kernel of this map is the ideal $(x^2)$, and $\phi$ is surjective, so $K[x]/(x^2)$ is isomorphic to $K\dot{+}K\bar{x}$ by the First Isomorphism Theorem.