Because $f(1, 1) = 1$, we start by considering the base case for the function $f$ when $n=1$ and $m$ is any positive integer:

For $m=1$, we have $f(1, 1) = 1$.

Using the second condition, we recursively find $f(m, 1)$:
$$f(m+1, 1) = 2f(m, 1) = 2(2f(m-1, 1)) = \cdots = 2^m \cdot f(1, 1) = 2^m.$$

This allows us to establish that:
$$f(m,1)=2^{m-1}.$$

Now, using the first condition, we can iteratively calculate $f(m, n)$ where $n > 1$:
$$f(m, n+1) = f(m, n) + 2.$$

Repeatedly applying this condition starting from $f(m, 1)$ for any specific $m$, we get:
$$f(m, 2) = f(m, 1) + 2 = 2^{m-1} + 2, \\ f(m, 3) = f(m, 2) + 2 = 2^{m-1} + 4, \\ \vdots \\ f(m, n) = 2^{m-1} + 2(n-1).$$

Thus, in general, we can represent $f(m, n)$ as:
$$f(m, n) = 2^{m-1} + 2(n-1).$$

Applying this formula to find $f(2010, 2011)$, we obtain:
$$f(2010, 2011) = 2^{2010-1} + 2(2011-1) = 2^{2009} + 2 \cdot 2010.$$

Since $2^{2009}$ is equal to $2^{2010} \div 2$, we can write:
$$f(2010, 2011) = \frac{2^{2010}}{2} + 4020 = 2^{2009} + 4020.$$

But we need this value in terms of $2^{2010}$ to match the answer choices, so multiplying both terms in the sum by 2 we get:
$$f(2010, 2011) = 2^{2010} + 2 \cdot 4020 = 2^{2010} + \boxed{ 4020 + 2 }$$