The second expression isn't really what's intended - in most situations in algebra, infinite sums don't make sense (one can consider Laurent series however). The correct expression is $K(\alpha)=\{f(\alpha)\mid f\in K[x]\}$ or equivalently $K(\alpha)=\{\textstyle\sum_{i=0}^\infty k_i\alpha^i\mid k_i\in K, \text{ and almost all }k_i=0\}.$ To show that the two definitions are equivalent, you just need to show that $L=\{f(\alpha)\mid f\in K[x]\}$ is the smallest field containing $K$ and $\alpha$. This amounts to showing that $L$ is a field, and that any field containing both $K$ and $\alpha$ will necessarily contain $L$.
It should be fairly clear that $L$ is a ring - we have $0$ and $1$, the sum of $f(\alpha)$ and $g(\alpha)$ is just $(f+g)(\alpha)$ where we certainly have $f+g\in K[x]$, and similarly with multiplication, etc. In fact, we can consider the evaluation homomorphism $\text{ev}_\alpha:K[x]\to L$ defined by $\text{ev}_\alpha(f)=f(\alpha).$ This is a surjective ring homomorphism, with kernel $J=\ker(\text{ev}_\alpha)=\{f\in K[x]\mid f(\alpha)=0\}.$ By the first isomorphism theorem, we have that $K[x]/J\cong L.$ By the assumption that $\alpha$ is algebraic, the kernel $J$ is non-trivial, because we know there is some non-zero polynomial with $\alpha$ as a root. In fact, $J=(m_\alpha)$ where $m_\alpha\in K[x]$ is the minimal polynomial for $\alpha$. We know that the minimal polynomial is irreducible as an element of $K[x]$, and thus $J$ is a prime ideal (because $K[x]$ is a UFD). In a PID, a non-zero prime ideal is maximal. Thus, $L\cong K[x]/J$ is a field.
Finally, note that any field containing $K$ and $\alpha$, being closed under addition and multiplication, has to contain $\alpha^n$ for all $n\in\mathbb{N}$, and therefore also contain $k\alpha^n$ for all $k\in K$ and $n\in\mathbb{N}$, and therefore also contain $k_0+k_1\alpha+\cdots+k_n\alpha^n$ for all $k_i\in K$ and $n\in\mathbb{N}$. These are the elements of $L$, so any field cotnaining $K$ and $\alpha$ must also contain $L$.