See, here is the conundrum of introductory aerodynamics/fluid mechanics books. Lift is hard. There is simply no simple way of explaining lift. Why would there be? It is only fair that you need quite some math to figure out the pressure distribution, which is the pressure field around an arbitrary body in airflow, and as you can imagine, that is no easy task. Who gets to say that just because lift is essential to flight, it must be readily understandable?
First off $V_{\mathrm{free stream}}\ne V_1\ne V_2$, and $P_{\mathrm{atmosphere}}\ne P_1\ne P_2$, so here is your answer.
Secondly in a more accurate analysis $V_1$, $V_2$, $P_1$, $P_2$ can not be assumed constant at either under or over the wing.
A more accurate portrayal of lift is most easily achieved by simplifying the airflow first to 2D-potential flow, i.e. $\exists \varphi, v=\nabla\varphi$ then assuming $\rho=\text{const}$, then we would get $\nabla^2\varphi=0$ the Laplace equation. The Bernoulli's equation is used here to link $p$ singlesidedly to $v$ hence $\varphi$, if you apply Bernoulli's equation right from the beginning, how do you even know $V_1>V_2$ without making it an assertion?
(A ramification of making the flow a potential flow is we automatically set $T=\text{const}$ as well.)
Now we have only a single variable, namely $v$, and there are multiple interesting ways to solve the Laplace equation. But perhaps the way that provides the most insight into lift generation is through a conformal mapping, i.e. $f\colon\mathbb{C}\mapsto\mathbb{C}$ analytical and $f^\prime\ne0$. Conformal $f$ has the property that if $\nabla^2\varphi=0$ then $\nabla^2(\varphi\circ f)=0$ and $\nabla^2(\varphi\circ f^{-1})=0$.
As you can see where this is going, we study the potential flow $\varphi_0$ around a cylinder then find a $f$ that maps the cylinder to the wing profile and $\varphi=\varphi_0\circ f$ automatically yields the potential flow around the wing.
There are three kinds of basic solution to the flow around the cylinder: rectilinear, vortex and doublet. And per the linearity of the Laplace equation, any superposition of the three solutions is also a solution. Similarly, any flow around a wing can be seen as superpositioning of three respective $\varphi_0\circ f$.
Below is the visualization of the solution of the pressure field of flow around a cylinder (red, purple = high pressure, green, blue = low pressure, free stream flows from right to left):
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Note that just like I said, there are three basic solutions, the pressure shown above is the result of changing the coefficient of combining the solutions.
Here is the airfoil version, note the high degree of similarity
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A most useful result of this approach is a direct proof of Kutta–Joukowski theorem. This theorem states that the lift, defined as the component of net force acting on a body immersed in rectilinear airflow that is perpendicular to the free stream's velocity vector, of an arbitrarily shaped body in an inviscid potential flow is given by
$$L=\rho V_{\infty}\Gamma,$$ where $\Gamma=\int_C v\,\mathrm{d}s$ for any route $C$ that surrounds the body. This tool, not the Bernoulli's principle, is the real workhorse of aerodynamicists.
Speaking of potential flow and Bernoulli's equation, here is an interesting fact:
From the differential form of the momentum equation
$$\rho\dfrac{\mathrm{D}}{\mathrm{D}t}v+\nabla p=0$$
with the assumption that $\rho=\text{const}$, $\dfrac{\partial(\cdot)}{\partial t}=0$ (steady flow) we get
$$\rho\dfrac{\mathrm{D}}{\mathrm{D}t}v+\nabla p=\rho(v\cdot\nabla)v+\nabla p=\nabla\left(\rho\dfrac{v\cdot v}{2}+p\right)=0,$$
which suggests the total head $$H=\rho\dfrac{V^2}{2}+P$$ is constant not only on streamline, but everywhere in the whole domain! This is a stronger version of Bernoulli's Law, implicit in the Newton's second law.
Please do notice I have only mentioned lift under the ideal circumstance of inviscid potential flow, and the solution given by this theory deviates from real life in a significant way. For example, you can tell from experience that there is no way a cylinder can stand in water flow and not feel any drag, yet the solution to flow around cylinder says so. This is called d'alembert paradox. The answer to this paradox is viscosity of water. The viscosity of water prevents a full pressure recovery on the rear half of the cylinder, and the flow would separate near the top and bottom of the cylinder. Viscosity is also important for airfoil, firstly because it is the primary reason why wings stall, secondly, it has an intricate relation to exactly which solution will be established around the wing, and this determines through K-L law the lift, i.e. frictional drag actually dictates pressure drag and lift!!!
EDIT: The lift equation is $L=\frac{1}{2}\rho C_LAv^2$. As you would have guessed it, the circulation $\Gamma$ is proportional to airspeed. But the reason why increasing airspeed increases circulation is even more arcane and too long for one answer.
EDIT: The lift equation is derived from the K-L law mentioned above. $C_L$ is defined as $\frac{L}{1/2\rho v^2 A}$, not obtained theoretically from the airfoil shape and $A$, as the equation would have you erroneously believe.
