Note that the remainder when $2^n$ is divided by $3$ is $1$ when $n$ is even, and $2$ when $n$ is odd.
Hence $\left[\frac{2^n}{3}\right] = \frac{2^n-1}{3}$ when $n$ is even, and $\left[\frac{2^n}{3}\right] = \frac{2^n-2}{3}$ when $n$ is odd. It follows that
$$\begin{align*} S &= \left[\frac{1}{3}\right] + \left[\frac{2}{3}\right] + \left[\frac{2^2}{3}\right] + \dots + \left[\frac{2^{2014}}{3}\right] \\ &= 0 + \left(\frac{2}{3} - \frac{2}{3} + \frac{2^2}{3} - \frac{2}{3}\right) + \left(\frac{2^3}{3} - \frac{2}{3} + \frac{2^4}{3} - \frac{2}{3}\right) + \dots + \left(\frac{2^{2013}}{3} - \frac{2}{3} + \frac{2^{2014}}{3} - \frac{2}{3}\right) \\ &= \left(\frac{2}{3} + \frac{2^2}{3} - 1\right) + \left(\frac{2^3}{3} + \frac{2^4}{3} - 1\right) + \dots + \left(\frac{2^{2013}}{3} + \frac{2^{2014}}{3} - 1\right) \\ &= \left(\frac{2}{3} + \frac{2^2}{3} + \frac{2^3}{3} + \dots + \frac{2^{2014}}{3}\right) - 1007 \\ &= \frac{2^{2015} - 2}{3} - 1007 \end{align*}$$
The last two digits of powers of $2$ are listed as follows:
02, 04, 08, 16, 32, 64, 28, 56, 12, 24, 48, 96, 92, 84, 68, 36, 72, 44, 88, 76, 52, 04, 08, ...
The pattern repeats when the exponent is increased by $20$. So the last two digits of $2^{2015}$ are the same as those of $2^{15}$, i.e. $68$.
Now write $2^{2015} - 2 = 100k + 66$. Since $\frac{2^{2015}-2}{3}$ is an integer, $k$ is a multiple of $3$, and so we write $k = 3m$. Thus the last two digits of $S$ are the same as those of $100m + 22 - 7$, i.e. $15$.