Best way to deal with surge in current from DC motors
Is thee some standardized way to solve this problem?


Yes: use a capacitor.


  but is it all?


Yes.


  I can use a power supply to replace the battery and measure the actual peak current and its duration, but what to do after that?


You could estimate the necessary capacity. (I use \$U\$ for voltage, not \$V\$)

$$C={Q \over U}$$

So the question is how many Q \$Q\$ are needed for the spike. For current you know that

$$
\begin{align}
I &amp;={dQ \over dt}\\
Idt &amp;= dQ\\
\underset{spike}{\int} Idt &amp;= Q
\end{align}
$$

Basically speaking, the area under the spike tells you how many \$Q\$s you need. Of course the battery can handle some baseline current, which means a certain amount of \$Q\$s per time. The spike on top of that must be handled by the capacitor.

This is a rough estimate. Not every capacitor can take arbitrary high current spikes. Check the datasheets. Now get a capacitor with more than the estimated capacity and do your tests again. Scope the current from the battery and the current from the capacitor. The spike on the battery current should be reduced as the current for the spike comes from the capacitor.

Then put some of those capacitors in parallel to your battery. How many depends on how often a spike occurs and how fast the capacitors are recharged. 



from the comments:


  This battery has its own built in switch: won't it cause harm to have a very large capacitor in parallel to the battery, when i turn it on? 
  If the capacitor is large enough to sustain the spike I mentioned, won't it cause an even larger one onto the battery, when it's activated?
  I thought I should add some small resistor in series to the capacitor. But then that too should be taken into account when calculating the value of the capacitor, right?


It depends on how you place the resistor. 


If you place it between battery and capacitor


it's not "getting in the way" when the capacitor unloads to balance the spike,
but it's always "in the way" of the battery. All current from the battery goes through the resistor, which means a permanent power loss.

If you place it in series to the capacitor


it is not in the way of the current going to the motor
but it's in the way of the current coming from and going to the capacitor