First you need to understand why we attach something called a shunt to convert a galvanometer into an ammeter after all? This is because a galvanometer is a very sensitive device. It gives a full scale deflection (maximum possible reading) for current of the order of microamperes (normally current encountered is at least of the order of milliamperes), so if you connect it directly in a circuit, the needle will reach the maximum value and start oscillating rapidly (often with a characteristic sound). Secondly a galvanometer has an appreciable resistance (~100ohms) which changes the current that was initially in circuit.
Let's assume that the current in a circuit is I just before the point of attachment of the galvanometer and the shunt. At the junction, the current will fork into two: I' and i where I' is the current through the shunt and i is the current through the galvanometer.
What you need to understand is that while doing any conversion, the value of the shunt is something we have to calculate for a particular maximum value of current we require in the circuit. What's known is only the resistance of the galvanometer and a quantity called it's figure of merit (figure of merit is just the current required for a unit deflection). What we're doing is basically calibrating the value of the current through the galvanometer such that it gives us an idea of the current through the circuit. (In your case this is what enables it to measure a current greater than what actually passes through it. It=galvanometer)
In questions you'll be given a value of current which produces a maximum deflection in the galvanometer or the figure of merit of the galvanometer and the number of divisions a readings runs into.
Getting back, it follows from Kirchhoff's junction rule that the current at the point where it 'forks off' would be I= I' + i. In other words the current in the circuit split into two at the junction. A gross oversimplification would be something like this: Charges? They're like us driving on a road. We prefer to take the least clogged lanes. Much like us, charges will take the path which is the least clogged, i.e. the path which offers the least resistance. Here that path is through the shunt resistance.
So that does it for the stuff that happens at the junction. Now since the shunt and galvanometer are connected in parallel, the voltage drop across both would be the same. Mathematically, I'S= iG where S and G are the shunt and galvanometer resistance respectively.
Also, I'= I - i
Which makes us arrive at the formula, S= iG÷ (I - i)
Having rambled on for a few paragraphs, I realise that that might not have answered your doubt. I'll try to rephrase. Connecting two resistances in parallel we achieve a net resistance which is smaller than the smallest value of resistance connected in the combination. That's the case with the present scenario. We want a very small resistance introduced into the circuit when we add the galvanometer and that's why we add the shunt. Next, by the traffic analogy, the current which is flowing through the galvanometer is much smaller than what flows through the shunt (current which flows through the shunt, can be said to be very close to the value of the current which flows through the circuit). The formula establishes a relationship between the current which flows through both and calibrates the galvanometer to read the current through the circuit (current through the circuit is very very very close to the current through the shunt!) And this is how we're able to measure a value of current which is much greater than what actually passes through the galvanometer
Note: Though I haven't said much about the figure of merit, it's pretty much what the entire conversion revolves around.
Hope that helps.