I am studying some stuffs relative to PDEs, and appear the following sentences that for are very artificial and I do not understand.
By considering the relation on $x=a$ $$-A (\partial_x f - \partial_x g) = \partial_t f - \partial_t g + h, $$ by decomposing the relation before into the direction $(\partial_x f - \partial_x g)$ and its orthogonal direction, we conclude that $$ A=-\frac{(\partial_x f - \partial_x g)}{|(\partial_x f - \partial_x g)|^2} \left( \partial_t f - \partial_t g + h \right) \left|_{x=a}\right. $$ and $$ (\partial_x f - \partial_x g)^{\bot} \left( \partial_t f - \partial_t g + h \right) \left|_{x=a}\right. =0. $$
How is possible deduce the last two equalities ? Which means the last two equalities ? which method are considering to obtain it ?
$\require{amsmath}$
Call $\pmb{\phi} = \pmb{f} - \pmb{g}$, it is important to realize here that $\pmb{\phi}$ is a vector quantity, so let's go back to your original equation
$$ -A\partial_x\pmb{\phi} = \partial_t \pmb{\phi} + \pmb{h} \tag{1} $$
Evaluate $\partial_x \pmb{\phi}$ at $x=a$, remember that is a vector, you can multiply (inner product) both sides of equation (1) by that vector
$$ -A(\partial_x \pmb{\phi}) \cdot (\partial_x \pmb{\phi}) = (\partial_x \pmb{\phi})\cdot(\partial_t \pmb{\phi} + \pmb{h})= -A |\partial_x \pmb{\phi}|^2 \tag{2} $$
from here you can get $A$
$$ A = -\frac{\partial_x \pmb{\phi}}{|\partial_x\pmb{\phi}|^2}\cdot (\partial_t\pmb{\phi} + \pmb{h}) \tag{3} $$
Let's go back to the statement that $\partial_x \pmb{\phi}$ is a vector, as such you can define a normal vector to it, call it $(\partial_x \pmb{\phi})^\perp$, multiply both sides of equation (1) by this quantity
$$ -A \underbrace{(\partial_x \pmb{\phi})\cdot (\partial_x\pmb{\phi})^\perp}_{=0, ~~\text{because they are perpendicular}} = (\partial_t \pmb{\phi} + h) \cdot (\partial_x \pmb{\phi})^\perp = 0 $$