Winternitz-OTS+ with Poly1305
The requirement of the function in Winternitz OTS+ is a pseudorandom function family, or PRF: a family of functions $F_K\colon \{0,1\}^n \to \{0,1\}^m$ such that if $K$ is uniformly distributed in all $k$-bit strings, then $F_K$ is indistinguishable from a uniform random choice of function $U\colon \{0,1\}^n \to \{0,1\}^m$.  Specifically, the best known random algorithm $A(G)$ taking a function $G\colon \{0,1\}^n \to \{0,1\}^m$ to a true/false value and having a plausible computational cost has a negligible PRF advantage, $$\operatorname{Adv}^F_{\mathrm{PRF}}(A) = |\Pr[A(F_K) = 1] - \Pr[A(G) = 1]|,$$ which compares the probability that $A$ returns true for $F_K$ under uniform random distribution on $K$ to the probability that $A$ returns true for a uniform random function $U$.

Poly1305 is a one-time message authentication code, or one-time authenticator/MAC: a family of functions $H_K\colon \{0,1\}^n \to \{0,1\}^t$ such that if $K$ is uniform random, then given up to a single message $M$ and its authentication tag $T = H_K(M)$, an adversary has no hope of finding another message $M'$ and authentication tag $T'$ such that $T' = H_K(M')$—it resists forgery after a single message.  That is, the forgery probability for a random algorithm $A(M, T)$ giving $(M', T')$, namely $\Pr[M' \ne M, H_K(M') = T']$, is negligible.

A PRF makes a passable MAC, even a one-time MAC, although as you note Poly1305 is a much more efficient one-time MAC than a PRF.  However, a one-time MAC does not make a passable PRF.  Why not?  The adversary's algorithm $A$ in the notion of PRF advantage is allowed to call the function more than once.

To make this concrete, consider the secure one-time MAC $H_K(M)$ for $K = (K_0, K_1) \in \operatorname{GF}(2^{128}) \times \operatorname{GF}(2^{128})$ and $M \in \operatorname{GF}(2^{128})$ defined by $$H_{K_0, K_1}(M) = K_0 M + K_1.$$  (This is essentially GHASH, of AES-GCM, limited to one message block, which is a polynomial evaluation MAC much like Poly1305.  We can naturally interpret any 128-bit string as an element of $\operatorname{GF}(2^{128})$ given a choice of irreducible degree-128 binary polynomial like the standard GHASH one.)

How do we define the algorithm $A$ to distinguish $H_K$ from $U$?  We can first compute $T = K_0 M + K_1$ and $T' = K_0 M' + K_1$ by evaluating the putative $H_K$ on two different inputs, and then solve for $K_0$ and $K_1$ using secondary school algebra.  Then we can immediately distinguish $H_K$ from a uniform random function $U$ with high probability by computing the candidate function on $M''$ and checking whether we get $K_0 M'' + K_1$: with high probability, a uniform random function will give something else.

Actually, Winternitz OTS+ requires a slightly weaker notion than pseudorandom function family, namely key one-wayness: for uniform random key $K$ and input $X$, letting $Y = F_K(X)$, the success probability of the best known algorithm $A(X, Y)$ for recovering the key (or an equivalent), $$\operatorname{Adv}^F_{\mathrm{KOW}}(A) = \Pr[Y = F_{A(X, Y)}(X)]$$ is negligible, as long as it has a plausible computational cost, where $Y = F_K(X)$.  I leave it as an exercise to the reader to define such an algorithm $A$ for a polynomial evaluation MAC like the above example.