Actually, for any such projective $g$ and any $E$ there exists an $f$ such that this fails. In fact you may even assume that $f$ is a double cover. (I suspect that it also fails without the projective assumption, but this seems convincing enough).

Let $g:Y\to Z$ be a projective morphism, $Y$ smooth, $Z$ quasi-projective and $E$ a coherent sheaf on $Y$. Assume that $R^ig_*(E\otimes N)=0$ for all $N$ line bundles and $i>1$. Then there exists a finite, flat $f:X\to Y$ such that $X$ is smooth and $H^1(X,f^*E)\neq 0$. Notice that this is less than assumed and probably even less is enough for the above claim.

Claim Let $L$ be a $g$-ample line bundle on $Y$. Then $g_*(E\otimes L^{-m}) = 0$, but $Rg_*(E\otimes L^{-m})\neq 0$ for $m\gg 0$.
(Here $Rg_*$ stands for the total push-forward and this statement is equivalent to saying that there exists an $i$ such that $R^ig_*(E\otimes L^{-m})\neq 0$).

Proof The first claim is obvious, we can "kill" every section in $E$ by "dividing" with sections of $L$ enough times. The second one is relatively easy using Grothendieck duality and observing that $g_*(F\otimes L^m)\neq 0$ for $m\gg 0$ and any $F$ (and you just have to use an appropriate $F$ that comes from GD). I am sure this can be proved by alternative means. $\square$

So, now take $N=L^m$ such that $g_*(E\otimes N^{-1}) = 0$ and $Rg_*(E\otimes N^{-1})\neq 0$. By the assumption that $R^ig_*(E\otimes N)=0$ for all $i>1$ it follows that $R^1g_*(E\otimes N)\neq 0$. We may assume that $N^2$ is very ample and choose a general section $s\in H^0(Y,N^2)$. Let $\mathscr A=\mathscr O_Y\oplus N^{-1}$ and make it an $\mathscr O_Y$ algebra using $s$ to "wrap back" from $N^{-2}$ to $\mathscr O_Y$ as usual. Finally, let $X=\mathrm{Spec}_Y\mathscr A$ and $f$ the associated morphism. From the construction it is clear that $f$ is flat and finite of degree $2$. Again by the construction $f_*\mathscr O_X\simeq \mathscr A$ and then by the above, $R^1g_*(E\otimes f_*\mathscr O_X)\neq 0$.

Actually, for any such projective $g$ and any $E$ there exists an $f$ such that this fails. In fact you may even assume that $f$ is a double cover. (I suspect that it also fails without the projective assumption, but this seems convincing enough).

Let $g:Y\to Z$ be a projective morphism, $Y$ smooth, $Z$ quasi-projective and $E$ a coherent sheaf on $Y$. Assume that $R^ig_*(E\otimes N)=0$ for all $N$ line bundles and $i>1$. Then there exists a finite, flat $f:X\to Y$ such that $X$ is smooth and $H^1(X,f^*E)\neq 0$. Notice that this is less than assumed and probably even less is enough for the above claim.

Claim Let $L$ be a $g$-ample line bundle on $Y$. Then $g_*(E\otimes L^{-m}) = 0$, but $Rg_*(E\otimes L^{-m})\neq 0$ for $m\gg 0$.
(Here $Rg_*$ stands for the total push-forward and this statement is equivalent to saying that there exists an $i$ such that $R^ig_*(E\otimes L^{-m})\neq 0$).

Proof The first claim is obvious, we can "kill" every section in $E$ by "dividing" with sections of $L$ enough times. The second one is relatively easy using Grothendieck duality and observing that $g_*(F\otimes L^m)\neq 0$ for $m\gg 0$ and any $F$ (and you just have to use an appropriate $F$ that comes from GD). I am sure this can be proved by alternative means. $\square$

So, now take $N=L^m$ such that $g_*(E\otimes N^{-1}) = 0$ and $Rg_*(E\otimes N^{-1})\neq 0$. By the assumption that $R^ig_*(E\otimes N)=0$ for all $i>1$ it follows that $R^1g_*(E\otimes N^{-1})\neq 0$. We may assume that $N^2$ is very ample and choose a general section $s\in H^0(Y,N^2)$. Let $\mathscr A=\mathscr O_Y\oplus N^{-1}$ and make it an $\mathscr O_Y$ algebra using $s$ to "wrap back" from $N^{-2}$ to $\mathscr O_Y$ as usual. Finally, let $X=\mathrm{Spec}_Y\mathscr A$ and $f$ the associated morphism. From the construction it is clear that $f$ is flat and finite of degree $2$. Again by the construction $f_*\mathscr O_X\simeq \mathscr A$ and then by the above, $R^1g_*(E\otimes f_*\mathscr O_X)=R^1g_*E\oplus R^1g_*(E\otimes N^{-1})\neq 0$.