As current(DC) flows from positive to negative terminal which is actually flow of electrons which in turn flow from negative to positive terminal.
So the question is which side of battery starts conducting current first?
According to me it should be negative side as current is flow of electrons which originate from negative terminal.
Application of above concept is in following circuit where question is which bulb fuses first? 
[Physics] In which terminal does current starts to flow first i.e. positive or negative
chargeelectric-circuitselectric-currentelectricityelectrostatics
Related Solutions
Electric current is the rate of flow of electric charges across any cross-sectional area of a conductor. The direction of electric current is taken as the direction of flow of positive ions or opposite to the direction of flow of free electrons. Your assumption is not necessary here... Electrons always flow from negative terminal to positive terminal.
$$i=\frac{dq}{dt}$$
When current flows through an electrolytic solution or during the process of electrolysis, The plate towards which positive ions (cations) flow is called the cathode and the plate towards which negative ions (anions) flow is called the anode.
Wikipedia says clearly,
In an electrochemical cell, The electrode at which electrons leave the cell and oxidation occurs is called anode and the electrode at which electrons enter the cell and reduction occurs is called cathode. Each electrode may become either the anode or the cathode depending on the direction of current through the cell. A bipolar electrode is an electrode that functions as the anode of one cell and the cathode of another cell.
So, the convention is totally based on our definition of the direction of current flow that it always flows opposite to the direction of electrons (i.e) electrons can be called as cations or anions depending on the usage. And based on this, we dump our thought that cathode should always be negative, etc...
I think this picture answers the question:
The battery's terminals are not charged. It's just a chemical reaction that starts the charge.
There's a $\text{Zn}$ and a $\text{CuSO}_4$. When they meet each other via a cable, the $\text{Zn}$ gives 2 electrons to $\text{CuSO}_4$ and becomes $\text{Zn}^{2+}$.
$\text{Cu}^{2+}$ of $\text{CuSO}_4$ becomes $\text{Cu}$ and becomes electrostatically neutral, and separates from $\text{SO}_4^{2-}$.
Then we have $\text{SO}_4$ with -2 charge and $\text{Zn}$ with +2 charge. So, $\text{SO}_4^{2-}$ moves to $\text{Zn}^{2+}$ to get bound to each other.
In the end, we have $\text{ZnSO}_4$ and $\text{Cu}$.
With two different batteries, it's not possible. The $\text{SO}_4^{2-}$ has nowhere to go and make new connections. Thus the chemical and electrical process doesn't start.
Best Answer
Propogational effects of the electrical signal are only meaningful when the wire has finite capacitance and inductance. If it does then the electric signal travels according to the telegrapher's equations with a speed given by $$u=\frac{1}{\sqrt{LC}}$$ neglecting resistance. Now coming to your question, it is not just the circuit that matters but also how the electric disturbances are introduced i.e. how the terminals are connected to the circuit. If both the terminals are connected simultaenously the disturbances will travel from both ends towards the bulbs (the voltage of each disturbance will be different of course, and differ by Voltage of the battery) and since the bulbs are symmetrically placed both will light up at the same time. In general which bulb will light up first depends on the position of the bulbs and the timing of the cpnnection of each terminal. Also I must emphasise that in realistic situations this will happen negligible timescales and even wothin those timescales it will be complicated and varying waveform that exists in the wire, not a simple travelling step.