Decomposition of $K$-finite functions

Question

It is a theorem that a representation $(\pi,V_\pi)$ of a compact group $K$ decomposes as a direct sum of irreducible representations. My question is about Deitmar's treatment (Automorphic Forms page 198) of $K$-finite functions $G\rightarrow\mathbb{C}$, for $G$ a locally compact group and $K$ a compact subgroup. These are the complex-valued functions $f$ such that the space of maps $x\mapsto f(k_1xk_2)$, as $k_1, k_2$ range over $K$ is finite dimensional. In example $7.5.26$ he gives the following decomposition. Consider the representation $\rho$ of the compact group $K\times K$ on $L^2(G)$ given by \begin{equation} \rho(k,l)\phi(x)=\phi(k^{-1}xl). \end{equation} Then the Hilbert space $L^2(G)$ has the decomposition \begin{equation} L^2(G)=\bigoplus_{\tau\in\widehat{K\times K}}L^2(G)(\tau) \end{equation} and we can write any $f\in C_c(G)$ (continuous, compactly supported function on $G$) as \begin{equation} f=\sum_{\tau\in\widehat{K\times K}}f_\tau. \end{equation} Now, Deitmar writes 'The function $f$ is $K$-finite if and only if this sum is finite i.e. if $f_\tau=0$ for almost all $\tau$. So the set of $K$-finite vectors is the algebraic direct sum of all isotypes $L^2(G)(\tau)$ for $\tau\in\widehat{K\times K}$'. My question is: if the sum is direct, why is the decomposition of $f$ not automatically finite? I thought elements of a direct sum were supposed to be nonzero for only finitely many components? Does this mean that the decomposition of a representation of a compact group in general allows vectors to have infinitely many nonzero components? Perhaps some distinction is made between 'algebraic direct sum' and 'direct sum' though I'm not clear on this.