Here is a heavy machinery proof for the vanishing $R^if_*V$. It uses the notion of coherent $\mathcal{O}_X$-modules and complex spaces, since we have to work on the fibres which may not be manifolds.
I'll try to formulate the answer in such a way that knowledge of coherent modules and complex spaces is not necessary in order to follow the logic as long as you trust me in everything I'm attaching big names to.
(Just pretend coherent module means vector bundle and complex space means complex manifold.)
Theorem (Andreotti-Grauert) — For any complex space $X$ and each coherent $\mathcal{O}_X$-module $F$ the sheaf cohomology group $H^i(X,F)$ vanishes as soon as $i>\dim(X)$.
(See, e.g., Corollary 4.15 in Demailly's Complex Analytic and Differential Geometry.)
Note that this is precisely the claim in case $Y$ is just a point.
Corollary — Let $f\colon X\to Y$ be a proper holomorphic map between reduced complex spaces and let $d$ be the maximal dimension the fibres of $f$ attain. Then $R^if_*(F) = 0$ for all coherent $\mathcal{O}_X$-modules $F$ and all $i>d$. More precisely, for each $y\in Y$, $(R^if_*(F))_y = 0$ for all $i>\dim(f^{-1}(y))$.
Proof.
It suffices to prove that the completion $(R^if_*F)_y^\wedge = (R^if_*F)_y\otimes_{\mathcal{O}_{Y,y}} \mathcal{O}_{Y,y}^\wedge$ vanishes for all $i>d$ since the completion $\mathcal{O}^\wedge_{Y,y}$ of $\mathcal{O}_{Y,y}$ with respect to the maximal ideal is faithfully flat.
By Grauert's comparison theorem (in the algebro-geometric world also known as the Theorem on formal functions), for each $y\in Y$, the completion of the stalk at $y$, $(R^qf_*F)_y^\wedge$, is isomorphic to a limit
$$(R^qf_*F)^\wedge_y\cong \varprojlim\nolimits_k H^i\left(f^{-1}(y),F_{(k)}\right)$$
for certain coherent sheaves $F_{(k)}$, $k\in\mathbb{N}$, on the fibre $f^{-1}(y)$. Thus, if $i>\dim(f^{-1}(y))$, then $(R^qf_*F)_y = 0$. Hence, if $i>d = \max_{y\in Y}\{\dim(f^{-1}(y))\}$, then $R^qf_*F = 0$, as claimed.
Thanks to tracing's comment below, we can conclude:
Corollary — Let $f\colon X\to Y$ be a proper holomorphic map between complex spaces and let $d$ be the maximal dimension the fibres of $f$ attain. Then $H^p(R^qf_*(F)) = 0$ for all coherent $\mathcal{O}_X$-modules $F$, all $i>d$ and all $q$.
Next, note that we can't replace 'maximal fibre dimension' by $\dim(X)-\dim(Y)$. Let $Y$ be any smooth surface and let $f\colon X\to Y$ be the blow up of some point $y\in Y$. That's clearly a proper surjective map of complex manifolds and $\dim(X)-\dim(Y) = 0$.
The sheaf $R^1f_*F$ is supported in $y$ and so $H^0(Y,R^1f_*F) = (R^1f_*F)_y$ for any coherent sheaf $F$ on $X$. Therefore, we only have to find a vector bundle on $Y$ such that $R^1f_*F \not=0$. Let's look at the line bundle $F := \mathcal{O}_Y(2E)$ where $E\subset Y$ is the exceptional divisor.
We'll use Grauert's comparison theorem, so I have to tell you what the $F_{(k)}$ are: The ideal sheaf of $E$ in $Y$ is $\mathcal{O}_X(-E)$ and so $F_{(k)} = F\otimes_{\mathcal{O}_X}\mathcal{O}_X/\mathcal{O}_X(-kE)$.
Note that $$\mathcal{O}_X(-kE)/\mathcal{O}_X(-(k+1)E) = \mathcal{O}_X/\mathcal{O}_X(-E)\otimes \mathcal{O}_X(-kE) = \mathcal{O}_E\otimes \mathcal{O}_X(-kE) = \mathcal{O}_E(k)$$
since the normal bundle of $E$ in $X$ is $\mathcal{O}_E(-1)$.
For the same reason, $F_{(1)} = F\otimes \mathcal{O}_{E} = \mathcal{O}_{E}(-2)$ and we get each $F_{(k)}$ sitting in the exact sequence
$$0\to \mathcal{O}_E(k-2) \to F_{(k+1)}\to F_{(k)}\to 0.$$
Since $H^1(E,\mathcal{O}_E(n))=0$ for all $n\geq -1$ this implies that $H^1(E,F_{(k+1)})\cong H^1(E,F_{(k)})$ for all $k\geq 1$; therefore, the limit in Grauert's comparison is constant:
$$(R^1f_*F)^\wedge_y \cong \varprojlim\nolimits_k H^1\left(E,F_{(k)}\right) = H^1(\mathbb{P}^1,\mathcal{O}(-2))\not=0$$
and so $R^1f_*F\not=0$.
This also shows that $H^0(Y,R^qf_*V)$ need not vanish for $q>\dim(X)-\dim(Y)$. Right now, I don't have a counter-example for $p>0$.
