The starting point of this question makes it hard to give a short, satisfying answer; maybe someone will come along and do a standout one with diagrams, but I don't have time for it. So I'll kick off with this:
With two resistors in series, there's more total resistance on the circuit, which means that less current flows. With less current flow, a given resistor drops less voltage because V=IR. Voltage drop isn't an intrinsic property of the resistor; resistance is.
UPDATE :
John : Thanks for data. Graph is ok. I note your intercept is E=3.94V but your calculations use E=4.5V. This explains the discrepancy in your results. If you use 3.94V you get r ranging from 1.59 to 1.76, close to slope value of 1.68 Ohms.
ORIGINAL ANSWER :
Your line of best fit gives an average internal resistance r based on all measurements. If data points do not lie exactly on this line then the value of r calculated for individual data points (measured pairs of V and I) will not be exactly the same as the slope of the line of best fit.
If you have drawn the line correctly some points will be above the line and some below, with about as many each side, and with the above and below points distributed randomly.
However, it sounds as though there is a consistent trend in your data points : eg all 'below' points at low current and all 'above' points at high current. This suggests that internal resistance was not in fact constant, within the limitations of experimental error. You do not say how big an effect this is : if small, you may be able to ignore it.
EMF and r should be measured when the current drawn is very small, ideally 0. Possibly you have taken readings at a high current, or you have taken a long time to take them. This can have two effects : (i) depleting the battery, reducing EMF, and (ii) increasing r because the battery is warming up and this increases internal resistance.
Your observation that internal resistance increased as current decreased suggests to me that you may have started readings with a high current then worked down to low current.
You will need to decide for yourself what went wrong, perhaps after consulting your teacher again and explaining how you took the readings.
Best Answer
There are three voltages in this circuit: the voltage, $V_{batt}$ across the battery terminals, the voltage, $V_r$ across the fixed resistor, $r$, and the voltage, $V_R$, across the variable resistor. By definition of voltage (aka potential difference), $$V_{batt}= V_r+V_R$$ To a fair approximation (provided that $r+R$ isn't too small), $V_{batt}$ is constant. The current in the circuit is $$I=\frac{V_{batt}}{R+r}$$ The pd across $R$ is therefore $$V_R=IR=\frac{V_{batt}R}{R+r}$$ It's easy to show that, for a given value of $r$, as $R$ is increased from zero, the value of $V_R$ increases towards a maximum value of $V_{batt}$.
It's the presence of $r$ that spoils your argument!