Ohm's Law gives me different values compared to the practical experiment with a multimeter - can you explain why?

I you really connected the ampmeter like your sketch you should be happy it was not destroyed. What it measures in this position is the current the batterie gives in a shortcut A ampmeter must always be in the circuit, so between the batterie and the resistors, Your ammeter is connected as a voltmeter.

As you are already aware, you should not have connected the ammeter across the battery terminals. Fortunately, you didn't destroy the meter. So, you may ask, what current were you measuring?

Since the input resistance of the ammeter is much less than the parallel combination of the three resistors, which is nominally 333 Ohms, that parallel resistance can be ignored. That means the current you were measuring is approximately the current that the internal battery emf (nominal 9 V) was delivering through the series combination of the internal resistance of the battery, $R_b$, and the input resistance of the ammeter, $R_{m}$ or

$$I=frac{9}{R_{b}+R_{m}}$$

Hope this helps.