A local homomorphism $f:A \longrightarrow B$ of local rings is unramified if $B/f(m_A)B$ is finite separable field extension of $A/m_A$, or, equivalently, if
(1) $f(m_A)B=m_B$, and $\\$
(2) the field $B/m_B$ is finite and separable over $A/m_A$.
However, a morphism $f:X \longrightarrow Y$, locally of finite-type is said to be unramified at $x$ if $m_x=m_yO_{X,x}$ and $O_{X,x}/m_x$ is a finite separable extension of $O_{Y,y}/m_{y}$.
It shouldn't be, according to what is written for a homomorphism:$m_x=f^{\sharp}_{x}(m_y)O_{X,x}$ and $O_{X,x}/f^{\sharp}_{x}(m_y)O_{X,x}$ is a finite separable extension of $O_{Y,y}/m_{y}$???
I'm confusing myself.
You are right, but your source it not wrong either. The two definitions that you present here just use different notation. If you have a morphism $f \colon A \rightarrow B$ of rings, then this morphisms turns $B$ into an $A$-algebra. If now $I \subset A$ is an ideal, it is common to denote by $IB$ the ideal generated by $I$ in $B$, where generated by $I$ in $B$ actually means that we consider $I$ as a subset of $B$ via $f$. To summarize, $IB$ is a notation for $f(I)B$.