This is probably better suited for https://math.stackexchange.com/ but it probably doesn't hurt to say a few things here given that it's a fundamental part of studying parabolic equations and geometric evolution equations such as the Ricci flow. It's also a prototype for other maximum principal arguments and the comparison principle for various weak forms of solutions to fully non-linear equations such as viscosity solutions.
The function $\varphi$ in fact solves the equation
$$
\partial_t \varphi = \Delta_{g(t)} \varphi + \langle X(t), \nabla \varphi\rangle + F(\varphi(t), t)
$$
since $\varphi$ depends on $t$ only and hence $\Delta_{g(t)} \varphi = 0$ and $\nabla \varphi = 0$ and has initial data satisfying $\sup_M u(\cdot, 0) \leq \varphi(0)$. Then the weak maximum principal says that this inequality is preserved for all $t>0$: $\sup_M u(\cdot, t) \leq \varphi(t)$.
To prove that the inequality is preserved, one typically proceeds by contradiction. First suppose that $\sup_M u(\cdot, 0) < \varphi(0)$. Consider the function
$$
v(x, t) = \varphi(t) - u(x, t).
$$
We have that $\inf_M v(x, 0) > 0$ and need to prove that $v(x, t) \geq 0$ for all $x \in M$ and $t \geq 0$. In fact, we can prove that $v(x, t) > 0$.
Now the contradiction: suppose not. There there will be first time $t_0$ and a point $x_0 \in M$ such that $v(x_0, t_0) = 0$. That is, $v(x, t) > 0$ for all $x \in M$ and all $t < t_0$ and $v(x, t_0) \geq 0$ for all $x \in M$. Thus $\partial_t|_{t=t_0} v(x_0, t) \leq 0$ and $x_0$ is a spatial minimum of the function $x \mapsto v(x, t_0)$. Then we also have $\Delta_{g(t_0)} v|_{x=x_0} \geq 0$ and $\nabla v|_{x=x_0} = 0$.
Therefore at $(x_0, t_0)$, we have
$$
\tag{1}\label{eq:min}
\partial_t v - \Delta_{g(t)} v - \langle X(t), \nabla v \rangle \leq 0.
$$
On the hand, since $u$ is a sub-solution, $\varphi$ is a solution and everything (except perhaps $F$) is linear:
$$
\begin{split}
\partial_t v - \Delta_{g(t)} v + \langle X(t), \nabla v \rangle &= \partial_t \varphi - \left(\partial_t u - \Delta_{g(t)} u - \langle X(t, \nabla u \rangle\right) \\
&\geq F(\varphi(t_0), t_0) - F(u(x_0, t_0), t_0) = 0
\end{split}
\tag{2}\label{eq:solns}
$$
since $\varphi(t_0) = u(x_0, t_0)$.
Equation \eqref{eq:solns} does not quite contradict equation $\eqref{eq:min}$ because both could in fact be $0$. To obtain the contradiction, consider something like $v_{\epsilon} (x, t) = \varphi(t) - u(x, t) + \epsilon e^t$ for $\epsilon > 0$ and then send $\epsilon \to 0$.
This new $v_{\epsilon}$ will also allow you to get the result with only weak inequality: $\sup_M u(\cdot, 0) \leq \varphi(0)$. Essentially that is a continuity argument.
Another way to approach the contradiction argument, which works already with weak inequality (but you'll still need to add $\epsilon e^t$ to get a contradiction), is to suppose the result is false, so that there exists a $T > 0$ such that
$$
\inf_{x \in M, t \in [0, T]} \varphi(t) - u(x, t) < 0.
$$
Then choose $x_0 \in M$ and $t_0 \in [0, T]$ to realise the $\inf$ and the same argument works. Some messing around is also necessary here to deal with the $F$ term when you do it this way. The way around is to add $\epsilon e^{\lambda t}$ for suitably chosen $\lambda$.
This should be in many books dealing with PDE such as Evans. Another source for maximum principles in particular is Protter and Weinberger.
