Working of potentiometer 
So can you simply explain why current increases in lower loop as we move towards A why it increases in upper loop as we move towards X and please give explanation based upon what changes occur voltage between A and X. Also why there is 1V and -4V potential at terminals.



 A: Presumably there is a resistive wire between A and X (implied by the question).
The two batteries have their positive terminals connected at point P - I am going to assume that part of the wire has no resistance.
This gives us the following picture:

As the contact point (let's call it C, not shown explicitly in the diagram) moves closer to A, the resistance of the wire that is supporting the 1V potential difference gets smaller, and the current in the ammeter will increase. As we move C closer to B, there will be more resistance in the wire. Note that this is independent of the voltage across the upper circuit - we can ignore this because we have no information about resistance in the lower circuit and assume it to be zero.
Here is a more formal analysis of the situation:

I am assuming the only resistance in the circuit is provided by the potentiometer wire, where $R_1+R_2=R$, some constant value (which is not given, but which we don’t need to know). This means that we can write $R_2 = R - R_1$ which leaves us with just one variable related to the position of the potentiometer.
Now according to Kirchoff’s Law, we can write the currents in the circuit as the sum of current $I_1$ in the upper loop, and $I_2$ in the lower loop. Because the voltage drop around each of the loops must be zero, it then follows that
$$V_1 - I_1 (R-R_1) - (I_1+I_2) R_1 = 0\\
V_2 - (I_1+I_2)R_1 = 0$$
From the second of these equations it follows that $(I_1+I_2)R_1 = V_2$ , the voltage at point $C$. Substituting that into the first equation we find
$$V_1 - I_1(R-R_1) - V_2 = 0$$
which we can solve for $I_1$:
$$I_1 = \frac{V_1-V_2}{R-R_1}$$
In other words, as the slider moves further to the right, the current $I_1$ is increasing. You can easily see that this is so because point $C$ is fixed at -1 V, so with a constant voltage difference between B and C, and a decreasing resistance $R_2 = R-R_1$, the current $I_1$ must get bigger.
We can equally solve these equations for $I_2$, by substituting for $I_1$ in the second equation:
$$V_2 - (\frac{V_1-V_2}{{R - R_1}+I_2)R_1} = 0\\
 I_2 = \frac{V_1-V_2}{R-R_1} - \frac{V_2}{R_1}$$
Because $V_1 = 2 V_2$ this equation becomes nicely symmetrical:
$$\begin{align}
 I_2 &= \frac{V_2}{R-R_1} - \frac{V_2}{R_1}\\
&= \frac{V_2 (2R_1 - R)}{R_1(R-R_1)}\end{align}$$
This shows that the current $I_2$ will be zero when the slider is exactly in the middle - that is, when $R_1 = R - R_1$. Again, this is easily explained: if you consider the slider to be disconnected, then the voltage at the mid point would be exactly - 1 V; connecting a slider that also has a potential of -1 V will not cause a current to flow. Moving the slider away from the center will cause the (absolute value of the) current to increase - and as you can see, the sign changes as you go through zero (the midpoint of the wire).
