What follows is too long to fit in a comment.

Some thoughts on the second problem. Let us consider the problem for $t$ taking values in the compact interval $[0,T]$. The general case perhaps can be approximated by this case. If the $t$ variable were discrete and finite the problem would be: we have a sequence of functions $f_1 \ge f_2 \ge f_3 \cdots \ge f_n$ (let us make the simplifying assumption that $f_i$ are all positive; otherwise, replace $f_i$ with $f_i +C$ where $C$ is $ C = -\min_{i,x} f_i(x)$), can one find convex and increasing $\{h_i\}$ and $\{g_i\}$ such that $f_i = h_i - g_i$? The answer to this question is obviously yes. For example, $h_i = \sum_{j\le i} f_i$ and $g_i = \sum_{j < i} f_i$ is one possible solution. The problem with this solution when $t$ is continuous is that the $h_i$ and $g_i$ would explode as the discretization of $t$ is refined. Thus, one needs to choose $h$ and $g$ in a way that they increase slowly. How slowly can this increase be? We can start with $h_1 = f_1$ and $g_1 = 0$. $h_2$ and $g_2$ will be of the following form: $$ h_2 = h_1 + S_2 = f_1 + S_2, $$ and $$ g_2 = g_1 + R_2 = R_2 $$ such that: 1) $S_2, R_2 \ge 0$, 2) $h_2 = f_1 + S_2$, $g_2 = R_2$ are convex and 3) $h_2 - g_2 = f_1 + S_2 - R_2 = f_2$. The last of these is equivalent to $R_2 = f_1 - f_2 + S_2$. Thus, what we are looking for is a function $S_2$ satisfying the above conditions. There will be many such $S_2$, the goal is to choose $S_2$ in a minimal way so that $h_i$ and $g_i$ grow slowly.The best would be: $S_2 = 0$, which is indeed a solution when $f_1 - f_i$ is convex. If $f_0 - f_t$ is convex for all $t \in [0,T]$ then this solution directly generalizes to the original continuous time problem.

Now, let us consider the case when $f(t,x)$ is such that $\frac{\partial^2}{\partial x^2}f(t,x)$ is continuous in $(t,x)$ and $x$ also takes values in a compact set $K$. The following type of argument quickly comes to mind. Define $$\tau \doteq \{t: \exists g, h:[0,t]\times K\rightarrow {\mathbb R}, f = h-g, h,g \text{ convex in } x \text{ and increasing in } t \}.$$ $0$ is clearly a member of this set. One can perhaps argue that $\sup \tau$ must be $T$ as follows. If $t_0 = \sup \tau < T$ then one can slightly modify $h(t_0,x)$ and $g(t_0,x)$ to obtain functions $h(t_0+\delta,x)$ and $g(t_0+\delta,x)$ whose difference will be $f(t_0+\delta,x)$ [the slight modification is made possible by the fact that the second derivative of $f$ with respect to $x$ barely changes between $t_0$ and $t_0 + \delta$].

What follows is too long to fit in a comment.

Some thoughts on the second problem. Let us consider the problem for $t$ taking values in the compact interval $[0,T]$. The general case perhaps can be approximated by this case. If the $t$ variable were discrete and finite the problem would be: we have a sequence of functions $f_1 \ge f_2 \ge f_3 \cdots \ge f_n$ (let us make the simplifying assumption that $f_i$ are all positive; otherwise, replace $f_i$ with $f_i +C$ where $C$ is $ C = -\min_{i,x} f_i(x)$), can one find convex and increasing $\{h_i\}$ and $\{g_i\}$ such that $f_i = h_i - g_i$? The answer to this question is obviously yes. For example, $h_i = \sum_{j\le i} f_i$ and $g_i = \sum_{j < i} f_i$ is one possible solution. The problem with this solution when $t$ is continuous is that the $h_i$ and $g_i$ would explode as the discretization of $t$ is refined. Thus, one needs to choose $h$ and $g$ in a way that they increase slowly. How slowly can this increase be? We can start with $h_1 = f_1$ and $g_1 = 0$. $h_2$ and $g_2$ will be of the following form: $$ h_2 = h_1 + S_2 = f_1 + S_2, $$ and $$ g_2 = g_1 + R_2 = R_2 $$ such that: 1) $S_2, R_2 \ge 0$, 2) $h_2 = f_1 + S_2$, $g_2 = R_2$ are convex and 3) $h_2 - g_2 = f_1 + S_2 - R_2 = f_2$. The last of these is equivalent to $R_2 = f_1 - f_2 + S_2$. Thus, what we are looking for is a function $S_2$ satisfying the above conditions. There will be many such $S_2$, the goal is to choose $S_2$ in a minimal way so that $h_i$ and $g_i$ grow slowly.The best would be: $S_2 = 0$, which is indeed a solution when $f_1 - f_i$ is convex. If $f_0 - f_t$ is convex for all $t \in [0,T]$ then this solution directly generalizes to the original continuous time problem (i.e., if $f_0-f_t$ is convex then $h(t,x) = f(0,x)$ and $g(t,x) = f(0,x) - f(t,x)$ is a solution).

Now, let us consider the case when $f(t,x)$ is such that $\frac{\partial^2}{\partial x^2}f(t,x)$ is continuous in $(t,x)$ and $x$ also takes values in a compact set $K$. The following type of argument quickly comes to mind. Define $$\tau \doteq \{t: \exists g, h:[0,t]\times K\rightarrow {\mathbb R}, f = h-g, h,g \text{ convex in } x \text{ and increasing in } t \}.$$ $0$ is clearly a member of this set. One can perhaps argue that $\sup \tau$ must be $T$ as follows. If $t_0 = \sup \tau < T$ then one can slightly modify $h(t_0,x)$ and $g(t_0,x)$ to obtain functions $h(t_0+\delta,x)$ and $g(t_0+\delta,x)$ whose difference will be $f(t_0+\delta,x)$ [the slight modification is made possible by the fact that the second derivative of $f$ with respect to $x$ barely changes between $t_0$ and $t_0 + \delta$].