How are the sterile neutrinos $\nu_s$ different from the heavy right-handed fields $N_R$ (or $\nu_R$)? To generate three light neutrino mass eigenstates via type-I seesaw, we include a set of $n$ (need not be equal to 3) heavy$^1$ right-handed fields $N_R$ in addition to three $\nu_L$ fields to the Standard Model. Then type-I seesaw mechanism gives 3 light and $n$ heavy mass eigenstates which are linear combinations if $\nu_L$ and $N_R$ fields. 
The $N_R$ fields are gauge singlets, and therefore do not have any Standard Model (SM) interactions. Sterile neutrinos $$ also do not have SM interactions. This reference says $\nu_s$ fields are sterile neutrinos without defining it or distinguishing it from $N_R$ (or $\nu_R$) fields. While this reference says $\nu_R$ are sterile without talking about $\nu_s$ notation at all.
$\bullet$ Then how are the $N_R$ fields different from sterile neutrinos?
$\bullet$ Existing answers claim that $N_R$ is same as $\nu_s$. Indeed both are defined to be having no SM interactions. But we know that sterile neutrinos $\nu_s$ taking part in oscillation with active neutrinos must be light. On the other hand, the $N_R$ fields must be heavy for type-I seesaw to work.

$^1$ Heavy in the sense that $M$ in the term $M\bar{N_R^c}N_R$ is around the GUT scale. But of course $M_R$ aren't mass eigenstates.
 A: They are not different. There is only a lot of different (and sometimes confusing) terminology on the market. For instance, in the case of Dirac neutrinos, you also need the right-handed neutrinos (for the mass term, for instance). Some people will call these right-handed components sterile, some won't.  
UPDATE: There can be arbitrarily many right-handed (or sterile) neutrinos, see e.g. this paper and references therein. A priori, they can be light or heavy, and they may or may not mix with the active neutrinos. The notion "right-handed" refers to the chirality of a state that carries the same lepton number as an active neutrino particle (as opposed to antiparticle). Of course, if lepton number is broken, this notion can be confusing. It might therefore be better to distinguish between "active neutrinos" and "singlet neutrinos". But the bottom-line, or the answer to your question, is that right-handed and sterile neutrinos are not different.     
A: Before spontaneous symmetry breaking, we have the lepton doublets $l_L$ (which contain the left-handed neutrinos), and the neutral singlets $N_R$. For the latter, we have $D_\mu N_R = \partial_\mu N_R$, so they don't have interactions with the gauge fields (which usually come from $\bar{\psi}\gamma^\mu D_\mu\psi$). This is the reason they're called sterile, they can't interact electromagnetically or through the weak or the strong forces. However, nothing forbids them to have Yukawa interactions $\sim \bar{l}_L \phi N_R$ and Majorana masses $\sim \bar{N}_R^c N_R$, so this terms should be included in the Lagrangian.
After spontaneous symmetry breaking, $l_L$ splits into the electron and neutrino $\nu_L$. There's a non-diagonal mass matrix for $(\nu_L, N_R^c)$. The fields $(\nu_{\text{light}}, \nu_{\text{heavy}})$ that diagonalize the mass matrix are linear combinations of the ones that come from the components of the gauge eigenstates. Therefore, we can't talk about them being sterile because they are not singlets, as they don't even have well-defined transformation laws under the gauge group.
A: Theoretically speaking they are the same field, but when one refers to sterile neutrino typically small masses and suppressed mixing angles are adopted. When this sterile neutrino is heavy, with masses above 1 MeV,
the term right-handed neutrino is used more often.
See Ref. PhysRevD.105.035016, or arXiv:2108.11398.
Look through the paragraph related to Eq. (2).
I hope I have helped you.
