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If $D\in\mathrm{Pic}\\, X$, in fact if $D$ is any sheaf on $X$ and $f:X\to Y$ is a continuous map, then for any $V\subseteq Y$ open $H^0(V,f_*D):=H^0(f^{-1}V,D)$ by definition, so in particular $H^0(Y,f_*D):=H^0(X,D)$ and you don't need any of the assumptions.

Am I missing something???

[EDIT] In view of the latest edit to the question this is no longer an answer. However, I will not delete it as I think it contributed to the revision of the question, so it should be part of the record.

If $D\in\mathrm{Pic}\\, X$, in fact if $D$ is any sheaf on $X$ and $f:X\to Y$ is a continuous map, then for any $V\subseteq Y$ open $H^0(V,f_*D):=H^0(f^{-1}V,D)$ by definition, so in particular $H^0(Y,f_*D):=H^0(X,D)$ and you don't need any of the assumptions.

If $D\in\mathrm{Pic}\\, X$, in fact if $D$ is any sheaf on $X$ and $f:X\to Y$ is a continuous map, then for any $V\subseteq Y$ open $H^0(V,f_*D):=H^0(f^{-1}V,D)$ by definition, so in particular $H^0(Y,f_*D):=H^0(X,D)$ and you don't need any of the assumptions. Am I missing something???

If $D\in\mathrm{Pic}\\, X$, in fact if $D$ is any sheaf on $X$ and $f:X\to Y$ is a continuous map, then for any $V\subseteq Y$ open $H^0(V,f_*D):=H^0(f^{-1}V,D)$ by definition, so in particular $H^0(Y,f_*D):=H^0(X,D)$ and you don't need any of the assumptions. 

Am I missing something???

If $D\in\mathrm{Pic}\\, X$ and $f:X\to Y$ is a morphism, then for any $V\subseteq Y$ open $H^0(V,f_*D):=H^0(f^{-1}V,D)$ by definition, so in particular $H^0(Y,f_*D):=H^0(X,D)$ and you don't need any of the assumptions. Am I missing something???

If $D\in\mathrm{Pic}\\, X$, in fact if $D$ is any sheaf on $X$ and $f:X\to Y$ is a continuous map, then for any $V\subseteq Y$ open $H^0(V,f_*D):=H^0(f^{-1}V,D)$ by definition, so in particular $H^0(Y,f_*D):=H^0(X,D)$ and you don't need any of the assumptions. Am I missing something???