You're on the right track. I would view things as follows.
Let $Q=f(q,p)$ and $p=g(q,Q)$. Then, we have both
$$dQ=f_1(q,p)dq+f_2(q,p)dp \tag 1$$
and
$$dp=g_1(q,Q)dq+g_2(q,Q)dQ \tag 2$$
Substitution of $dQ$ as given in $(1)$ into $(2)$ yields
$$\begin{align}
dp&=g_1(q,Q)dq+g_2(q,Q)\left(f_1(q,p)dq+f_2(q,p)dp \right)\\\\
&=\left(g_1(q,Q)+g_2(q,Q)f_1(q,p)\right)dq +f_2(q,p)g_2(q,Q)\,dp\tag 3
\end{align}$$
Inasmuch as the differential terms on both sides of $(3)$ must be equal, we obtain
$$\bbox[5px,border:2px solid #C0A000]{f_2(q,p)g_2(q,Q)=\frac{\partial Q}{\partial p}\times \frac{\partial p}{\partial Q}=1}$$
which was to be shown along with
$$g_1(q,Q)+g_2(q,Q)f_1(q,p)=\frac{\partial p}{\partial q}+\frac{\partial p}{\partial Q}\times \frac{\partial Q}{\partial q}=0$$
I think the confusion comes from the commonly used abuse of notation in calculus namely the abusive usage of free and bound variables.
Here we are dealing with operations acting on functions. Strictly speaking functions are things that take objects and output obects. In formal theory such as ZF, they can be defined as sets of ordered pairs for which there are no two pairs that are equal on their first element but differ on the second.
It's common to specify a function by its value on an arbitrary element of it's domain like $ x \mapsto x^2 $. Here x is a bound variable it has no independent meaning by itself it just acts as a dummy to specify any object in the domain. Now since that's often too verbose for the purposes of computation usually the $\mapsto$ is skipped and the function's are specified like $x^2$ which is now ambiguous since $x$ is now a free variable and could be just some number, but that's ok since people usually understand what's intended by such notation (namely that this variable is actually bound and not free).
The derivatives follow similar convention. $\frac{df(x)}{dx}$ is understood as the function $x \mapsto \lim\limits_{h\rightarrow 0} \frac{f(x+h) - f(x) }{h}$.
However this slight abuse goes even further beyond. Namely nonsensical expressions such as $\frac{df}{dx}$ are used, where now we have two free variables that have no connection whatsoever and it requires full knowledge of the context to understand what the author of such nonsense has intended. What's even more confusing can be the use of the same symbols as both functions and numbers in different contexts, with nonsense like $x=x(y)$, $y=y(x)$, $\frac{dy}{dx}$ where we might've used $x$ and $y$ as both fixed numbers, functions and dummy variables just one sentence ago.
For example when we write stupid things like $\frac{df(g(x))}{dx} = \frac{df}{dg} \frac{dg}{dx}$ what are abbreviating $\frac{df(g(x))}{dx} = \frac{df(g(x))}{dg(x)} \frac{dg(x)}{dx}$ which is again another abuse that stands for: Take the function $f$, take it's derivative $f^\prime$, take the function $g$, take it's derivative $g^\prime$, now compose $f^\prime \circ g$ and form a function which is the product of the two, that's the map
$x \mapsto (f^\prime \circ g)(x) g^\prime(x)$.
Now knowing with all that let's tackle your case.
By the nonsense $\frac{\partial g(t, a(t), b(t))}{\partial t} = \frac{\partial g}{ \partial t} + \frac{\partial g}{ \partial a} \frac{da}{dt} +\frac{\partial g}{ \partial b} \frac{db}{dt}$ and more specially the last term $\frac{\partial g}{ \partial b} \frac{db}{dt}$ that bothers you, we mean the following.
By $\frac{\partial g}{ \partial b}$ we mean take the function $g$, form it's partial derivative with respect to the third component (the number of the argument is the only thing that you need in order to take the partial derivative all the $x$'s and $t$'s are just abused dummies) namely $\partial_3 g$, now superpose that with the function $b$, hence $(x_0, x_1, x_2) \mapsto (\partial_3 g) (x_0, x_1, b(x_2))$.
What's $\partial_3 g$ ? Well in our abusive notation it's just
$$(\alpha,\beta,\gamma) \mapsto \frac{\partial g(\alpha,\beta,\gamma)}{\partial \gamma} = \frac{d}{d\gamma} \int_\beta^\gamma f(t, \alpha) dt$$
You should recognize this as just the normal fundamental theorem of calculus. $\gamma$ is just a bound variable it doesn't "depend" on the bound dummy variable of integration, it doesn't depend on $\alpha$ or anything else.
I've used $\alpha$ $\beta$ and $\gamma$ because they are dummies and I can use whatever I want. We could've used the more familiar
$$(t,a,x) \mapsto \frac{\partial g(t,a,x)}{\partial x} = \frac{d}{dx} \int_a^x f(y, t) dy = f(x,t)$$
So now we need to compose our map with the function $b$, and so we get that the map
$$(t,a,x) \mapsto f(b(x), t)$$
If you evaluate this at $(t,t,t)$ ( now $t$ is free here) it would be $f(b(t),t)$. The important thing is to recognize that these variables are bound (if you've not evaluated the function at some point) and have no meaning on their own.
So yes this is the standard fundamental theorem of calculus there's no trickery involved it's just that the notation is ambiguous and confusing.
Also while it's extremely tedious you can also use the more general abstract notation for the chain rule $D_{\mathbf{a}}(f \circ g) = D_{g(\mathbf{a})}f \circ D_{\mathbf{a}}g $, and avoid any abuse of bound variables by introducing new functions all the time when it's appropriate, forming compositions $\circ$ and using the $\mapsto$ notation which will avoid all ambiguity.
But like I said since this is really tedious for doing actual computations on concrete functions, the non-abusive notation is never used outside of very theoretically oriented texts.
I also think that it's useful to read this
https://en.wikipedia.org/wiki/Bound_variable .
Best Answer
I found the same Poisson's equation/problem in this article:
Apparently $$u(x, y) = y(1 - y) x^3$$
Lets check that's real true, given the conditions in my question. Lets first check the boundary conditions.
$u(0, y) = y(1 - y) * 0^3 = 0$
$u(x, 0) = 0(1 - 0) x^3 = 0*1 * x^3 = 0$
$u(x, 1) = 1 *(1 - 1) * x^3 = 1 * 0 * x^3 = 0$
$u(1, y) = y(1 - y) * 1 = y(1 - y)$
So the boundary conditions are satisfied.
Now lets check the second order partial derivatives of $u(x, y)$ with both respect to $x$ and $y$.
$$u_x = \frac{\partial u(x, y)}{\partial x} = y(1 - y) 3x^2$$
then
$$u_{xx} = \frac{\partial^2 u(x, y)}{\partial x^2} = \frac{\partial u_x}{\partial x} = y(1 - y) 6x$$
Now lets take with respect to $y$
$$u_y = \frac{\partial u(x, y)}{\partial y} = x^3 - 2yx^3$$
$$u_{yy} = \frac{\partial^2 u(x, y)}{\partial y^2} = \frac{\partial u_y}{\partial x} = - 2x^3$$
Now lets verify that $\frac{\partial^2u(x, y)}{ \partial x^2} + \frac{\partial^2 u(x, y)}{ \partial y^2} = f(x, y) $.
So
$$6yx - 6xy^2 + (-2x^3) = 6yx - 6xy^2 -2x^3 = f(x, y)$$