On the derivation of the backward Kolmogorov equation I started reading this article that I found on the internet about the backward Kolmogorov equation http://apghosh.public.iastate.edu/files/EORMS_BFeqnsDiff_Feb1-2010.pdf
I selected the relevant passages here



It seems to me that the equation 
$$u(t+ h,x) = \int P(h,x,dy)u(t,y) $$
is wrong, 
that is, if we put 12 and 8 of the text togeter shouldn't we obtain
$$u(t,x) = \int P(h,x,dy)u(t +h,y)  \tag 1$$
or 
$$u(t- h,x) = \int P(h,x,dy)u(t,y) \tag 2$$
If we proceed with the expression in (1) we get that
\begin{align*}
\frac{u(t+h, x) - u(t,x)}{h}  &= \frac{\int P(h,x,dy)[u(t+h,x)-u(t+h,y)]}{h}\\
& \sim  \frac{1}{h} \int P(h,x,dy)\bigg[\frac{\partial u (t+h,x)}{\partial x}(x-y) + \frac{1}{2}\frac{\partial^2 u (t+h,x)}{\partial^2 x}(x-y)^2\bigg]\\
&\xrightarrow[h \to 0]{} -\frac{\partial u (t+h,x)}{\partial x}b(x) + \frac{1}{2}\frac{\partial^2 u (t,x)}{\partial^2 x}\sigma(x)^2
\end{align*}
which gives
$$\partial_t u(t,x) = -\frac{\partial u (t+h,x)}{\partial x}b(x) + \frac{1}{2}\frac{\partial^2 u (t,x)}{\partial^2 x}\sigma(x)^2
 $$
whereas if we started from (2) we would get
\begin{align*}
\frac{u(t-h, x) - u(t,x)}{h}  &= \frac{\int P(h,x,dy)[u(t,y)-u(t,x)]}{h}\\
& \sim  \frac{1}{h} \int P(h,x,dy)\bigg[\frac{\partial u (t,x)}{\partial x}(y-x) + \frac{1}{2}\frac{\partial^2 u (t,x)}{\partial^2 x}(y-x)^2\bigg]\\
&\xrightarrow[h \to 0]{} \frac{\partial u (t,x)}{\partial x}b(x) + \frac{1}{2}\frac{\partial^2 u (t,x)}{\partial^2 x}\sigma(x)^2
\end{align*}
which gives
$$-\partial_t u(t,x) = \frac{\partial u (t+h,x)}{\partial x}b(x) + \frac{1}{2}\frac{\partial^2 u (t,x)}{\partial^2 x}\sigma(x)^2
 $$
Or more concisely
$$ \partial_t u(t,x) + A u(t,x) = 0 $$
The questions therefore are
1) is there indeed a typo on the text?
2) if so, why do those different reasonings  yield different results, shouldn't they be the same?

I think you are right that the original equation is wrong.
In regard to your derivation based on (1),  I think you made a mistake. Basically, if $x$ is fixed, 
$f(x)-f(y)\approx -f'(x)(y-x)-\frac{1}{2}f''(x)(y-x)^2$
The second term also has a negative sign in front of it.
