The relevant security property of $\operatorname{Poly1305}_r$ is that it has bounded difference probability—that is, for any distinct messages $x \ne y$ of up to $L$ bytes, and any difference $\delta$, $$\Pr[\operatorname{Poly1305}_r(x) - \operatorname{Poly1305}_r(y) = \delta] \leq 8\lceil L/16\rceil/2^{106},$$ under random choice of $r$. (Here the subtraction is modulo $2^{128}$; internally, Poly1305 works modulo $2^{130} - 5$ and limits $r$ to $2^{106}$ possibilities to enable cheap arithmetic, which accounts for the weird constant factor $8/2^{106}$.)
A forger, given a legitimate message/authenticator pair $(m, a)$ related by $a = \operatorname{Poly1305}_r(m) + s$ for unknown $r$ and $s$, who tries find a forgery $(m', a')$ with $m' \ne m$ will be thwarted with high probability for any $m'$ and $a'$ because the one-time forgery probability is bounded by the difference probability:
\begin{align*}
\Pr&[a' = \operatorname{Poly1305}_r(m') + s
\mid a = \operatorname{Poly1305}_r(m) + s] \\
&= \Pr[a' = \operatorname{Poly1305}_r(m') + a - \operatorname{Poly1305}_r(m)] \\
&= \Pr[\operatorname{Poly1305}_r(m') - \operatorname{Poly1305}_r(m) = a' - a] \\
&\leq 8\lceil L/16\rceil/2^{106}.
\end{align*}
For NaCl crypto_secretbox_xsalsa20poly1305, the story essentially ends here—we derive an effectively independent $r$ and $s$ for each message by the PRF XSalsa20. For Poly1305-AES, the story also involves the Carter–Wegman method (paywall-free) of authenticating $n$ messages with independent random secrets $r, s_1, s_2, \dotsc, s_n$ using a universal hash family like Poly1305, and Shoup's instantiation with a block cipher like AES to derive $s_i = \operatorname{AES}_k(i)$ from a short key $k$ and unique message number $i$. (More background, history, and references.)
Why must $r$ be kept secret?
With $r$ an adversary could trivially forge authenticators. For example, given the one-time authenticator $a = \operatorname{Poly1305}_r(m) + s$ on the message $m$, the adversary could compute the one-time pad $s$ used to conceal the hash $\operatorname{Poly1305}_r(m)$ by $s = a - \operatorname{Poly1305}_r(m)$, and then—with full knowledge of the authenticator keys $r$ and $s$—forge the authenticator $a' = \operatorname{Poly1305}_r(m') + s$ for any $m' \ne m$. This attack works no matter how you pick $r$ and $s$, e.g. even if $s = \operatorname{AES}_k(i)$ for some AES key $k$ and message number $i$.