Here is what you should try: Consider the function $$ p(x,\lambda) = K_0(x,0,\ldots,0)+\lambda\ K_1(x,0,\ldots,0) + \cdots + \lambda^n\ K_n(x,0,\ldots,0) $$ For any analytic solution $\lambda=L(x)$ to this polynomial equation that is a simple root of $p(x,\lambda)=0$, then, by the Cauchy-Kovalevskya Theorem, there is a unique solution to your problem on an open neighborhood of $U\times \lbrace0\rbrace$ in $U\times I$ that satisfies $f_{tt}(x,0)=L(x)$. Thus, if there is more than one simple root, you will have more than one solution.
Added comment: Perhaps I should explain what is meant by 'simple root'. I mean that, first, $p\bigl(x,L(x)\bigr)\equiv0$ and, second, that, when one performs the polynomial division so as to get $p(x,\lambda) = \bigl(\lambda - L(x)\bigr)\ q(x,\lambda)$ where $q$ is analytic in $x$ and polynomial in $\lambda$, then $q\bigl(x,L(x)\bigr)$ is nonvanishing on $U$. Alternatively (i.e., a different way of saying the same thing), one could require that $p\bigl(x,L(x)\bigr)\equiv0$ while $p_\lambda\bigl(x,L(x)\bigr)$ is nonvanishing on $U$, where $p_\lambda$ means the partial derivative of $p$ with respect to $\lambda$.
Meanwhile, if you have an analytic solution $\lambda=L(x)$ to this polynomial equation that is not a simple root, there may still be a solution (or there may not), but there are more conditions, and you will need the theory of singular analytic PDE to check them. For example, see R. Gérard and H. Tahara, Singular nonlinear partial differential equations, Aspects of Mathematics, Friedr. Vieweg & Sohn, Braunschweig, 1996.
