Hint: Indeed, condition on $X(t)=n$. Then, note that given $X(t)=n$, the locations of the $n$ points are distributed as $n$ i.i.d. uniform$[0,t]$ rv's. This should lead you easily to the result, noting that ${\rm E}f(U)=\frac{1}{t}\int_0^t {f(w)dw} $, where $U$ is a uniform$[0,t]$ rv.
EDIT: More precisely (cf. cardinal's comment below), given $X(t)=n$, the points $W_1,\ldots,W_n$ are distributed as $n$ order statistics from the uniform distribution on $[0,t]$. However, the sum $f(W_1)+\cdots+f(W_n)$ (given $X(t)=n$) is equal in distribution to the sum $f(U_1)+\cdots+f(U_n)$ where the $U_i$ are i.i.d. uniform$[0,t]$ rv's, and thus the result follows straightforwardly using the hint.
EDIT (additional hint, in response to the OP's request): $ \sum\nolimits_{n = 0}^\infty {n{\rm P}(X(t) = n)} = {\rm E}[X(t)] = ? $ The left-hand side is given by $ \sum\nolimits_{n = 0}^\infty {n{\rm E}[f(U)]{\rm P}(X(t) = n)}, $ where $U$ is a uniform$[0,t]$ random variable.
EDIT (in response to the OP's request): In general, if $Y$ is a random variable with density function $h$, then ${\rm E}[f(Y)] = \int {f(y)h(y)dy} $. A uniform$[0,t]$ random variable, $U$, has constant density function $h(y)=1/t$ for $y \in [0,t]$ (and $0$ otherwise). From this it follows that ${\rm E}[f(U)] = \frac{1}{t}\int_0^t {f(y)dy} $.
As for the summation $\sum\nolimits_{n = 0}^\infty {n{\rm E}[f(U)]{\rm P}(X(t) = n)}$, first note that $ {\rm E}\Big[\sum\nolimits_{i = 1}^{X(t)} {f(W_i )} \Big] = \sum\limits_{n = 0}^\infty {{\rm E}\Big[} \sum\nolimits_{i = 1}^n {f(W_i )} |X(t) = n \Big]{\rm P}(X(t) = n). $ Now, recall that given $X(t)=n$, the sum $f(W_1)+\cdots+f(W_n)$ is equal in distribution to the sum $f(U_1)+\cdots+f(U_n)$, where the $U_i$ are i.i.d. uniform$[0,t]$ rv's. This accounts for the $n {\rm E}[f(U)]$ in the sum.
Further questions?