This is an exercise from Ravi Vakil's notes on spectral sequences.
Suppose you have a spectral sequence $E^{\bullet\,\bullet}_\bullet$ such that $E^{i\,j}_0$ is zero if either $i$ or $j$ is negative (so if you let your arrows point upward and rightward, the initial page is bounded below and to the left). Let $E^\bullet$ denote the total complex of $E^{\bullet\,\bullet}_\bullet$. Show that $H^0(E^\bullet) = E^{0\,0}_\infty = E^{0\,0}_2$, and show that this sequence is exact: $$ 0 \to E^{0\,1}_2 \to H^1(E^\bullet) \to E^{1\,0}_2 \xrightarrow{\partial^{1\,0}_2} E^{0\,2}_2 \to H^2(E^\bullet) $$
Showing that $H^0(E^\bullet) = E^{0\,0}_\infty = E^{0\,0}_2$ was pretty manageable, but trying to show that that sequence is exact got messy quickly. I wrote down what $E^{0\,1}_2$ and $H^1(E^\bullet)$ must be in terms of a bunch kernels and images, and tried to clean it up, but I couldn't manage to see what the natural map $E^{0\,1}_2 \to H^1(E^\bullet)$ must be. Should I just continue to dive into the messiness of kernels and images, or is there a slicker way of seeing what the maps are in this sequence? Also, notice this more general related question.
I would like to point out this is exercise 5.1.3 in Weibel's book (who does homology spectral sequences) ---I suppose the notes make reference to this. Of course it is not a bad idea to look at Weibel's book. Rotman has some nice notes on spectral sequences in his book too, and in particular treats the case of double complexes with care, as in the notes you link. I think his account is cleaner.
This can be settled looking at some steps of the spectral sequence. Say $\{E_r\}$ converges to $H$, so that $H^1$ and $H^2$ are the cohomology groups in your exact sequence. Recall that since $\{E_r\}$ converges to $H$, we have a filtration $F$ on $H$ and isomorphisms
$$F^p H^{p+q}/ F^{p-1}H^{p+q} \simeq E_\infty^{p,q}$$
This filtration is such that $F^{-1}H =0 $ and $F^n H^n = H^n$, a canonically bounded filtration.
Note that $E_{2}^{0,0} = E_\infty^{0,0}$ stabilizes since both differentials arriving and leaving $E_2^{0,0}$ are trivial. Now $E_\infty^{0,0} = F^0 H^0 / F^{-1} H^0 = H^0$ since $F^{-1}H^0 = 0$ and $F^0H^0 = H^0$ (the filtration is canonically bounded).
We now make some observations. First, by degree considerations as before, $E_2^{0,1} = E_\infty^{0,1} = F^0H^1$ injects canonically in $H^1$. This gives you the first arrow in the sequence.
The cokernel of this is $H^1/ F^0H^1 = E_\infty^{1,0}$. Again by degree considerations this is $E_3^{1,0}$ which is the kernel of $d_2^{1,0}$, so you can extend your sequence as stated.
The map $d_2^{0,1}$ maps to $E_2^{0,2}$ by definition, and the quotient of this by the image of $d_2^{0,1}$ is already $E_\infty^{0,2}$. This is $F^0 H^2$ which sits inside $H^2$, completing the sequence.