The answer is no, as the following example shows.
Let $Y$ be an elliptic curve, $X=Y \times Y$ and $f \colon X \to Y$ the projection onto the first factor.
Since $f$ has connected fibres, we have $f_* \mathcal{O}_X= \mathcal{O}_Y$. On the other hand, since $X$ is a product, for all $p \in Y$ we may identify canonically $H^1(f^{-1}(p), \mathcal{O}_{f^{-1}(p)})$ with $H^1(Y, \mathcal{O}_Y) \cong \mathbb{C}$, hence $R^1f_* \mathcal{O}_X=\mathcal{O}_Y$.
Then, if we apply the functor $f_*$ to the split exact sequence
$0 \to \mathcal{O}_X \to \mathcal{O}_X \oplus \mathcal{O}_X \to \mathcal{O}_X \to 0$,
we obtain
$0 \to \mathcal{O}_Y \to \mathcal{O}_Y \oplus \mathcal{O}_Y \to \mathcal{O}_Y\stackrel{\delta}{\to} \mathcal{O}_Y \to \mathcal{O}_Y \oplus \mathcal{O}_Y \to \mathcal{O}_Y$,
where $\delta$ is the zero map.