Moni Noar One Way Function: What does input is partioned mean? I am trying to understand An Implementation of Efficient Pseudo-Random Functions.

There, the following hash function is defined:


  We define $\hat{h}=\hat{h}_{r}: {\{0,1\}}^* \rightarrow {\bf Z}_R$ 
  as follows:
  $$ \hat{h}_r(m) = ({\sum_{i=1}^{k} m_i \cdot r^{i-1}}) \bmod R $$
  where $R$ is a 161-bit prime, $r$ (the key of $\hat{h}$) is a uniformly distributed 
  element in ${\bf Z}_{R}^*$ and the input $m$ is partitioned into a sequence,
  $\langle m_1 \dots m_k\rangle$, of elements in ${\bf Z}_{R}$.


I am having trouble understanding what "$m$ is partioned into a sequence,  of elements in $\textbf{Z}_R$ means. Does this mean we understand $m$ as a bitstring, iterate through it from left to right as long as the resulting number is $&lt;R$ and repeat this process until we covered the whole $m$ bitstring?
I am having trouble understanding what "$m$ is partioned into a sequence $\langle m_1, ..., m_k\rangle$ of elements in $\textbf{Z}_R$ means.


This is essentially what you said. Consider $n$ such that $\{0,1\}^n$, seen as integers, is a subset of $\textbf{Z}_R$ (e.g. $n = \lfloor\log_2R\rfloor$). Then regard $m$ as a long string $m = m_1\|\cdots\|m_N$ where each $m_i$ is in $\{0,1\}^n$ (if $n$ does not divide the bit-length of $m$ then apply a "good" padding to $m$, i.e. one that is prefix-free).

Just take into account that, as noticed by @poncho in the comments, this is not a PRF. This is a 2-universal hash function, and as shown in the post you cite, it is an ingredient for the construction of an actual PRF.