Why is the gate charge curve (Miller plateau) of MOSFETs dependent on Vds?
"Why is the Miller Plateau longer for bigger \$V_{\text{ds}}\$? "  

The short answer is that Miller Plateau width scales with the area under the curve for \$C_{\text{gd}}\$.  But why?  

What does the Miller Plateau show?  

The Miller effect exists because there is effective capacitance between the drain and gate of the FET (\$C_ {\text {gd}}\$), the so called Miller capacitance.  The curve of Figure 6 in the datasheet is generated by switching the FET on with a constant current into the gate, while the drain has been pulled up through a current limiting circuit to some voltage \$V_ {\text {dd}}\$.  After the gate voltage rises past the threshold and drain current reaches it limit (set by the current limiting circuit), \$V_ {\text {ds}}\$ starts to fall, displacing charge on \$C_ {\text {gd}}\$ through the gate.  While \$V_ {\text {ds}}\$ falls to zero volts, from \$V_ {\text {dd}}\$, \$V_G\$ is stuck by the displacement current from \$C_ {\text {gd}}\$ ... that' s the Miller Plateau.  

The Miller Plateau shows the amount of charge in \$C_ {\text {gd}}\$ by its width.  For a given FET the width of the Miller Plateau is a function of the voltage traversed by \$V_ {\text {ds}}\$ as it switches on.  The figure shows \$V_G\$ aligned with \$V_ {\text {ds}}\$ to make this clear.  

  

The gate charge curve for the IRFZ44 shows three spans of \$V_{\text{ds}}\$; Span1 is 0V to 11V, Span2 is 0V to 28V, and Span3 is 0V to 44V.  Now, some things should be clear:  


\$V_{\text{ds}}\$ Span3 > \$V_{\text{ds}}\$ Span2 > \$V_{\text{ds}}\$ Span1  
\$V_{\text{ds}}\$ Span3 includes Span2 and Span1.
\$C_{\text{gd}}\$ charge is greater for a larger \$V_{\text{ds}}\$ span.  
Miller Plateau will be wider with more \$C_{\text{gd}}\$ charge.  
More is more.  


Do these conclusions seem too hand wavy and snake oily to you?  Ok, then how about this?  

Why the Miller Plateau gets Wider for Higher \$V_{\text{ds}}\$ -- A Quantitative Look  

Start with the equation for charge on a capacitor:  

Q = CV      with a differential form    dQ = C dV  

Now \$C_{\text{gd}}\$ is not a constant, but some function of \$V_{\text{ds}}\$.  Looking at the curve in Figure 5 of the IRFZ44 data sheet for \$C_{\text{gd}}\$, we want some equation that is not infinity at zero \$V_{\text{ds}}\$ and falls off exponentially (ish).  I won't go into any details here about how this was done.  Just choose very simple forms that seem to match and try fitting them to the data.  So, not based on device physics, but just matches pretty good with pretty little effort.  Sometimes that's all that's required.  

\$C_{\text{gd}}\$ = \$\frac{C_{\text{gdo}}}{k_c \text{V}_{\text{ds}}+1}\$  

where
    \$C_{\text{gdo}}\$ = 1056 pF
    \$k_c\$ = 0.41  -- an arbitrary scaling coefficient  

Checking this fitted model to the datasheet we see:  

\begin{array}{ccc}
 V_{\text{ds}} &amp; C_{\text{gd}}\text{(data)} &amp; C_{\text{gd}}\text{(model)} \\
 \text{1V} &amp; 750pF &amp; 749pF \\
 \text{8V} &amp; 250pF &amp; 247pF \\
 \text{25V} &amp; 88pF &amp; 94pF
\end{array}  

So, after plugging the \$C_{\text{gd}}\$ model expression into the differential form of the charge equation, and integrating both sides we get:  

Q = \$\frac{C_{\text{gdo}} \log \left(k_c V_{\text{ds}}+1\right)}{k_c}\$ = \$\frac{\text{1056 pF } \log \left(\text{0.41 } V_{\text{ds}}+1\right)}{\text{0.41 }}\$  

A plot of Q shows that it always increases for larger changes of \$V_{\text{ds}}\$.  

  

The only way this would not be true would be if \$C_{\text{gd}}\$ became negative for some values of \$V_{\text{ds}}\$, which isn't physically realizable.  So, more is more.