The questions, in particular, are the two written in italics, but any other correction is welcome. Thank you in advance.
Let $(f,f^\#):(Y,\mathcal O_{Y})\to (X,\mathcal O_{X})$ be a morphism of LRS, so $f^\#:\mathcal O_X\to f_*(\mathcal O_{Y})$ is a morphism of sheaves; also, denote by $f^{\#}_{y}$ the composite $\mathcal O_{X,f(y)}\to f_*(\mathcal O_{Y})_{f(y)}\to\mathcal O_{Y,y}$ for every $y\in Y$. Letting $i:Y\to f(Y)$ be the inclusion, $f$ factorizes as $i\circ g$, where $g$ is induced in the obvious way, and $(f(Y),i^{-1}(\mathcal O_X))$ is a LRS. In general, do $i$ and $g$ extend to morphisms of LRS, respectively $(i,i^\#):(f(Y),i^{-1}(\mathcal O_X))\to (X,\mathcal O_{X})$ and $(g,g^\#):(Y,\mathcal O_{Y})\to (f(Y),i^{-1}(\mathcal O_X))$, such that $(i,i^\#)\circ (g,g^\#)=(f,f^\#)$?
General case. Call $\alpha:i^{-1}(f_*)\to g_*$ the (canonical) natural transformation of functors $\operatorname{Sh}(f(Y))\to \operatorname{Sh}(f(Y))$, and set $g^\#:=\alpha_{\mathcal O_Y}\circ i^{-1}(f^\#)$. Consider this commutative diagram, where (if not labelled) the vertical arrows are canonical isomorphisms, and the horizontal ones are the induced homomorphisms, existing in all the three cases, into $\mathcal O_{Y,y}$; since $f^\#_y$ (the lower row) is a local homomorphism, also $g^\#_y$ (the upper row) is. $\require{AMScd}$ $$\begin{CD} i^{-1}(\mathcal O_X)_{g(y)} @>>{g^\#_{g(y)}}> g_*(\mathcal O_Y)_{g(y)} @>>> \mathcal O_{Y,y} \\@AA{\operatorname{id}}A @AA{(\alpha_{\mathcal O_Y})_{g(y)}}A @AA{\operatorname{id}}A\\ i^{-1}(\mathcal O_X)_{g(y)} @>>{i^{-1}(f^\#)_{g(y)}}> i^{-1}(f_*(\mathcal O_Y))_{g(y)} @>>> \mathcal O_{Y,y} \\ @AAA @AAA @AA{\operatorname{id}}A\\ \mathcal O_{X,f(y)} @>>{f^\#_{f(y)}}> f_*(\mathcal O_Y)_{f(y)}@>>> \mathcal O_{Y,y} \\ \end{CD}$$
Hence $g$ extends to a morphism of LRS. Observe that there is a (canonical) natural transformation $\Phi:\operatorname{id}_{\operatorname{Sh}(X)}\to i_*(i^{-1})$ of functors $\operatorname{Sh}(X)\to \operatorname{Sh}(X)$; it gives raise to this commutative diagram, where the left column coincides with the (canonical) isomorphism $f_*(\mathcal O_Y)\to i_*(g_*(\mathcal O_Y))$. $\require{AMScd}$ $$\begin{CD} i_*(i^{-1}(\mathcal O_{X})) @>>{i_*(g^\#)}> i_*(g_*(\mathcal O_Y))\\@AA{\operatorname{id}}A @AA{i_*(\alpha_{\mathcal O_Y})}A \\ i_*(i^{-1}(\mathcal O_{X})) @>>{i_*(i^{-1}(f^\#))}> i_*(i^{-1}(f_*(\mathcal O_Y))\\ @AA{\Phi_{\mathcal O_X}}A @AA{\Phi_{f_*(\mathcal O_Y)}}A \\\mathcal O_{X}@>>{f^\#}> f_*(\mathcal O_Y)\\ \end{CD}$$
Setting $i^\#:=\Phi_{\mathcal O_X}$, the commutativity of the outer square means $(i,i^\#)\circ (g,g^\#)=(f,f^\#)$. It is left to prove that $i^\#_x$ is a local homomorphism for every $x\in f(Y)$. As $i$ is homeomorphic onto the image, the induced $i_*(\mathcal F)_{i(x)}\to\mathcal F_x$ is an isomorphism for every sheaf $\mathcal F$ on $f(Y)$; this holds, in particular, for $i_*(i^{-1}(\mathcal O_X))_{i(x)}\to (i^{-1}(\mathcal O_X)_x$. The composition (of isomorphisms) $$(\mathcal O_X)_{i(x)}\to (i^{-1}(\mathcal O_X))_x\to i_*(i^{-1}(\mathcal O_X))_{i(x)}$$ coincides with $(\Phi_{\mathcal O_X})_{i(x)}$, and $i^\#_x$ is $(\mathcal O_X)_{i(x)}\to (i^{-1}(\mathcal O_X))_x$ that is not only local, but isomorphic.
Open immmersions. If $(f,f^\#)$ is assumed to be an open immersion, one can define $g^\#$ analogously to above, and use that $f^\#_y$ is isomorphic to imply that $g^\#_y$ is. Plus, as $f$ is homeomorphic onto the image, $g$ is a homeomorphism.
Closed immersions. So far I'd say that any $(f,f^\#)$ factorizes through a morphism $(i,i^\#)$, whose underlying map is the inclusion, such that $i^\#_x$ is an isomorphism for every $x\in f(Y)$; moreover if $(f,f^\#)$ is an open immersion, the induced $(g,g^\#)$ is an isomorphism. If $(f,f^\#)$ is assumed to be a closed immersion instead, is it possible to factorize it also in a way that $(g,g^\#)$ is an isomorphism, and $i^\#_x$ surjective for every $x\in f(Y)$?
$(f(Y),i^{-1}(\mathcal O_X/\operatorname{ker}f^\#))$ is a LRS, and $f^\#$ induces an isomorphism of sheaves $F^\#:\mathcal O_X/\operatorname{ker}f^\#\to f_*(\mathcal O_Y)$ (the quotients are assumed to be sheaficated). Set $g^\#:=\alpha_{\mathcal O_Y}\circ i^{-1}(F^\#)$; then $g^\#_y$ is isomorphic because in this commutative diagram the lower row is isomorphic (as $f$ is homeomorphic onto the image). $\require{AMScd}$ $$\begin{CD} i^{-1}(\mathcal O_X/\operatorname{ker}f^\#)_{g(y)} @>>{g^\#_{g(y)}}> g_*(\mathcal O_Y)_{g(y)} @>>> \mathcal O_{Y,y} \\@AA{\operatorname{id}}A @AA{(\alpha_{\mathcal O_Y})_{g(y)}}A @AA{\operatorname{id}}A\\ i^{-1}(\mathcal O_X/\operatorname{ker}f^\#)_{g(y)} @>>{i^{-1}(F^\#)_{g(y)}}> i^{-1}(f_*(\mathcal O_Y))_{g(y)} @>>> \mathcal O_{Y,y} \\ @AAA @AAA @AA{\operatorname{id}}A\\ (\mathcal O_X/\operatorname{ker}f^\#)_{f(y)} @>>{F^\#_{f(y)}}> f_*(\mathcal O_Y)_{f(y)}@>>> \mathcal O_{Y,y} \\ \end{CD}$$
It is left to prove that $i$ extends to a morphism $(i,i^\#):(f(Y),i^{-1}(\mathcal O_X/\operatorname{ker}f^\#))\to (X,\mathcal O_X)$ such that $(i,i^\#)\circ (g,g^\#)=(f,f^\#)$. If $i^\#:=\Phi_{\mathcal O_X/\operatorname{ker}f^\#}\circ \pi$, it is equivalent to the commutativity of (the outer square in) this diagram. $$\begin{CD} i_*(i^{-1}(\mathcal O_X) @>>{i_*(i^{-1}(\pi))}> i_*(i^{-1}(\mathcal O_X/\operatorname{ker}f^\#) @>>{i_*(g^\#)}> i_*(g_*(\mathcal O_Y))\\@AA{\operatorname{id}}A @AA{\operatorname{id}}A @AA{i_*(\alpha)}A \\i_*(i^{-1}(\mathcal O_X) @>>{i_*(i^{-1}(\pi))}> i_*(i^{-1}(\mathcal O_X/\operatorname{ker}f^\#)) @>>{i_*(i^{-1}(F^\#))}> i_*(i^{-1}(f_*(\mathcal O_Y))\\ @AA{\Phi_{\mathcal O_X}}A @AA{\Phi_{\mathcal O_X/\operatorname{ker}f^\#}}A @AA{\Phi_{f_*(\mathcal O_Y)}}A \\\mathcal O_X @>>{\pi}> \mathcal O_X/\operatorname{ker}f^\# @>>{F^\#}> f_*(\mathcal O_Y)\\ \end{CD}$$
Similarly to before, for every $x\in f(Y)$, $i^\#_{x}$ is surjective as it coincides with the composition $$\mathcal O_{X,i(x)}\xrightarrow{\pi_{i(x)}}(\mathcal O_X/\operatorname{ker}f^\#)_{i(x)}\to i^{-1}(\mathcal O_X/\operatorname{ker}f^\#)_x.$$