How to find volume needed from a stock solution
This is calculated by equating the moles of the two solutions.

You are using $\pu{0.6402g}$ of sodium fumarate in $\pu{100mL}$ of solution. 

$$ \text{No. of moles} = \frac{\text{Given weight}}{\text{Molecular weight}} = \frac{0.6402}{160.04} = \pu{0.0040 moles} $$

$$ \text{The concentration of the solution can be found easily} = \frac{0.0040}{0.1} = \pu{0.04 M}$$
If you require a solution of concentration $40$mM, the stock solution that you have prepared is exactly that.
If you need a more diluted solution, you may use the formula:
$$ M_1 V_1 = M_2V_2$$
where, $M_1$ and $M_2$ are the initial and final concentrations, $V_1$ is the volume of stock you would require for the dilution, and $V_2$ is the volume of final desired concentration.
Let us assume you require $\pu{50 mL}$ of $\pu{10 mM}$ solution, and you have a stock of $\pu{100 mL}$ of $\pu{40 mM}$.
Then
$$
M_1 = \pu{40 mM}\\
V_1 = ? \\
M_2 = \pu{10 mM} \\
V_2 = \pu{50 mL}
$$
Plugging in the values, you will find out that you require $\pu{12.5 mL}$ of stock, and dilute it to $\pu{50mL}$ using the solvent, i.e., add $\pu{37.5 mL}$ of water to $\pu{12.5 mL}$ of stock solution.