Why is the momentum advection dependent on all 3 dimensions? For a 3D flow, the advection term in the x-momentum equation is $-\nabla \cdot (\rho u \vec{u})$. How does momentum flow in the y and z directions contribute to the momentum flux in the x direction?
 A: The point about this is that the velocity field not only specifies where something is moving but also also what is moving because momentum is proportional to velocity. Technically, momentum is similar to charge in electrodynamics, which is conserved because it moves along the velocity field of charges. However, since momentum is not only the "conserved charge" that gets transported, but also defines the velocity itself, the momentum flow tensor has this peculiar form.
This becomes apparent if you write the Cauchy momentum equation in integral form. Cauchy momentum equation in differential form is
$$\frac {\partial}{\partial t} (\rho\,\mathbf{u})
   + \nabla \cdot (\rho\,\mathbf{u} \otimes \mathbf{u})
 = \mathbf{f}$$
where $\mathbf{f}$ are all the interactions of a fluid element with its surroundings, e.g. pressure, gravity, etc.
If you integrate this over a fixed volume $V$ (Euler picture of fluid dynamics), you get (using Gauss' divergence theorem):
$$\frac {d}{d t} \int_V (\rho\,\mathbf{u}) dV
   + \oint_{\partial V}(\rho\,\mathbf{u}) (\mathbf{u}\cdot d\mathbf{A})
 = \int_V \mathbf{f} dV=:\mathbf{F}_V$$
The leftmost term is obviously the change of momentum, while the second term is the flow of momentum through the boundary of the volume $V$. Hence you can relate the change of momentum in the volume $V$ to changes due to the interactions with the surrounding fluid elements and the bare momentum flow that is transported through the boundary surface $\partial V$ of the volume. Let's denote momentum density as $\boldsymbol{\pi}:=\rho \mathbf{u}$ and total momentum in the volume $V$ by
$$\mathbf{P}_V:=\int_V (\rho\,\mathbf{u}) dV=\int_V \boldsymbol{\pi} dV$$
Then momentum conservation can be expressed as
$$\frac {d\mathbf{P}_V}{d t} =\mathbf{F}_V-\oint_{\partial V}\boldsymbol{\pi} (\mathbf{u}\cdot d\mathbf{A}) $$
Since $\mathbf{u}\cdot d\mathbf{A}$ is an inner product, that mainly depends on the angle between the particular boundary surface element and the flow velocity, it determines how much of the "charge" represented by the momentum density $\boldsymbol{\pi}$ will actually get transported through that element $d\mathbf{A}$ of the boundary surface. On the other hand, the presence of momentum density $\boldsymbol{\pi}$ in the integral on the right-hand side indicates, what gets transported (momentum!), and this includes, of course, all three components of momentum.
More specifically, and with reference to your question: in the most general case, $y$ and $z$ components of momentum density $\boldsymbol{\pi}$ are also going to be transported through a surface $dA_x$ oriented perpendicular to the $x$ coordinate direction.
