As described for example here, realized variance is the sum of squared returns - $\sum_{i=0}^{N-1} R_i^2$ in your notation.
Thus it is wrong to assume that the formula consists of:
- "normalizing factor" - $252$
- "variance of the logarithmic returns" in statistical terms - $\frac{1}{N}\sum_{i=0}^{N-1} R_i^2$.
Instead the two pieces of the formula are:
- realized variance - $\sum_{i=0}^{N-1} R_i^2$
- reciprocal of duration of time period - $\frac{252}{N}$
And since we are not talking about variance (in statistical sense) of the logarithmic returns, it does not make sense to ask whether and why logarithmic returns are supposed to have zero-expectation!
The realized variance is useful because it provides a relatively accurate measure of volatility of the underlying - $\sigma$.
The actual derivation of the formula is illustraded below.
Let $\alpha$ and $\sigma$ be constants, and define the geometric Brownian motion $$ S(t) = S(0) e^{\sigma W(t)+(\alpha-\frac{1}{2}\sigma^2)t}$$
Let $0 \leq T_1 < T_2$ be given and suppose we observe $S(t)$ for $T_1 \leq t \leq T_2$. Choose some partition of this interval $T_1 = t_0 < t_1 < \cdots < t_m = T_2$ and observe log returns $\log\frac{S_{t_{j+1}}}{S_{t_{j}}}$ over each of subintervals $[t_j, t_{j+1}]$:
$$\log\frac{S_{t_{j+1}}}{S_{t_{j}}} = \sigma (W(t_{j+1}) - W(t_j)) + (\alpha - \frac{1}{2}\sigma^2)(t_{j+1}-t_j)$$
The realized volatility $\sum_{j=0}^m\Big(\log\frac{S_{t_{j+1}}}{S_{t_{j}}}\Big)^2$ is:
$$\sum_{j=0}^m\Big(\log\frac{S_{t_{j+1}}}{S_{t_{j}}}\Big)^2 = \sigma^2 \sum_{j=0}^m \big(W(t_{j+1})-W(t_j)\big)^2 + \big(\alpha - \frac{1}{2}\sigma^2\big)^2 \sum_{j=0}^m (t_{j+1}-t_j)^2 + \\2\sigma(\alpha - \frac{1}{2}\sigma^2)\sum_{j=0}^m (W(t_{j+1})-W(t_j))(t_{j+1}-t_j)$$
Let $\|\Pi\| = \max_{j=0,1,\cdots m-1}(t_{j+1}-t_j)$. Then:
$$ \lim_{\|\Pi\|\to 0} \sum_{j=0}^m (t_{j+1}-t_j)^2 = 0 \\
\lim_{\|\Pi\|\to 0} \sum_{j=0}^m (W(t_{j+1})-W(t_j))(t_{j+1}-t_j) = 0 \\
\lim_{\|\Pi\|\to 0} \sum_{j=0}^m \big(W(t_{j+1})-W(t_j)\big)^2 = T_2-T_1$$
Thus:
$$\sigma^2 \approx \frac{1}{T_2-T_1}\sum_{j=0}^m\Big(\log\frac{S_{t_{j+1}}}{S_{t_{j}}}\Big)^2$$
Now you can see that $\frac{252}{N} = \frac{1}{T_2-T_1} $ and we don't need the assumption that logarithmic returns have zero-expectation. Instead it is assumed that $S(t)$ follows geometric Browninan motion with constant volatility.
The above derivation has been shamelessly stolen from here (3.4.3 Volatility of Geometric Brownian Motion)
Also have a look at "The Volatility Smile" by E. Derman. Chapter 4 "Variance Swaps" discusses how variance swaps can be replicated