Let $ (f,f^\flat)\colon (X,\mathscr O_X)\to (Y,\mathscr O_Y) $ be a morphism of ringed spaces (i.e., let $ f\colon X\to Y $ be continuous and let $ f^\flat\colon \mathscr O_Y\to f_*\mathscr O_X $ be a morphism of sheaves).
Let $ x\in X $. There are at least two ways to obtain a ring homomorphism $$ f_x\colon \mathscr O_{Y,f(x)}\to \mathscr O_{X,x}\text{.} $$
On one hand, we can say that $ f_x $ is the unique map making the diagram $$ \require{AMScd} \begin{CD} \mathscr O_Y(V) @>{({-})_{f(x)}}>> \mathscr O_{Y,f(x)}\\ @V{f_V^\flat}VV @VV{f_x}V\\ \mathscr O_X(f^*V) @>{({-})_x}>> \mathscr O_{X,x} \end{CD} $$ commutative for every open subset $ V\subset Y $ containing $ f(x) $, i.e., such that $$ f_x(t_{f(x)}) = f_V^\flat(t)_x $$ for every $ t\in \mathscr O_Y(V) $.
On the other, we can consider the adjoint sheaf map $ f^\flat\colon f^*\mathscr O_Y\to \mathscr O_X $ and take the stalk at $ x $ $$ f_x^\sharp\colon (f^*\mathscr O_Y)_x\to \mathscr O_{X,x} $$ and then compose $ f_x^\sharp $ with the canonical isomorphism $$ \mathscr O_{Y,f(x)} \cong (f^*\mathscr O_Y)_x $$ to get a map $ f_x\colon \mathscr O_{Y,f(x)}\to \mathscr O_{X,x} $.
I was trying to convince me that both the maps $ f_x $ that I have obtained are indeed the same, but I got stuck.
Is there an arrow nonsense way to prove this fact?