There is one deduction on page 14 I don't seem to be able to understand. It has been established that the increment of the functional can be expressed as:
$$\Delta J = \int_a^b \left[F\left(x, y+h, y^\prime + h^\prime\right) - F\left(x, y, y^\prime\right)\right] dx$$
This is supposedly equal to:
$$\int_a^b \left[F_y\left(x, y, y^\prime\right) h + F_{y^\prime} \left(x, y, y^\prime\right) h^\prime\right] dx + \cdots$$
Here, the dots "denote terms of order higher than 1 relative to $h$ and $h^\prime$" and this is supposed to (easily) follow from Taylor's theorem, but I don't see how. Can anyone explain this step to me?
Ah, misreading the text tripped me up. I edited in a missing prime, and with @jlammy's comment, everything makes sense now. We are evaluating $J[y] = \int_a^b F(x, y, y')\ dx$ at $J[y+h] = \int_a^b F(x, y+h, y'+h')\ dx$, and as far as the Taylor polynomial is concerned, we expand $T_1F$ around $(x,y,y')$. This gives:
\begin{align} T_1F &= F(x, y, y') \\ &\phantom{=} + \frac{\delta F}{\delta x}(x, y, y') (x-x) + \frac{\delta F}{\delta y}(x, y, y') (y+h-y) + \frac{\delta F}{\delta y'}(x, y, y') (y'+h'-y') \\ &= F(x, y, y') + \frac{\delta F}{\delta y}(x, y, y') h + \frac{\delta F}{\delta y'}(x, y, y') h' \end{align}
So $F(x, y+h, y'+h') = F(x, y, y') + \frac{\delta F}{\delta y}(x, y, y') h + \frac{\delta F}{\delta y'}(x, y, y') h' + \cdots$
Now,
\begin{align} \Delta J = J[y+h]-J[y] &= \int_a^b F(x, y+h, y'+h') - F(x, y, y')\ dx \\ &= \int_a^b F(x, y, y') + \frac{\delta F}{\delta y}(x, y, y') h + \frac{\delta F}{\delta y'}(x, y, y') h' - F(x, y, y') \ dx + \cdots \\ &= \int_a^b \frac{\delta F}{\delta y}(x, y, y') h + \frac{\delta F}{\delta y'}(x, y, y') h'\ dx + \cdots \end{align}