Continuous functions with convex level sets
Let's call the functions defined by Ali Taghavi to be sliced functions: a continuous function $\ f:\mathbb R^2\rightarrow\mathbb R\ $ is called sliced $\ \Leftarrow:\Rightarrow\ \ \forall_{c\in\mathbb R}\ f^{-1}(c)\ $ is convex.
NOTATION: $$\ [x;y]\ :=\ \{(1\!-\!t)\cdot x\ +\ t\cdot y\ :\ 0\le t\le 1\}\ $$ for arbitrary $\ x\ y\in \mathbb R^n\ $ and $\ n=0\ 1\ 2\ \ldots.\ $ Thus $\ [x;y]=[y;x]\ $ in every dimension including $\ n=1$.
THEOREM 0 Let $\ f:\mathbb R^2\rightarrow \mathbb R\ $ be sliced. Then $\ f^{-1}(C)\ $ is convex for every convex $\ C\subseteq\mathbb R$.
PROOF Let $\ C\subseteq \mathbb R\ $ be convex. Let $\ a\ b\in C.\ $ We want to show that
$$\ [a;b]\ \subseteq\ C $$
If $\ f(a)=f(b)\ $ then the above holds due to the convex part of the definition of a sliced function.
Now, assume $\ f(a)\ne f(b).\ $ Then $\ f^{-1}([f(a);f(b)])\ \cap\ [a;b]\ $ is a closed subset of $\ [a;b].\ $ Next, consider arbitrary $\ x\ y \in [a;b]\cap f^{-1}([f(a);f(b)],\ $ and $\ x\ne y.\ $ If $\ f(x)=f(y)\ $ then again
$$ [x;y]\ \subseteq f^{-1}(x)\ \subseteq\ f^{-1}([f(a);f(b)]) $$
And if $\ f(x)\ne f(y)\ $ then, due to continuity of $\ f\ $ (and of the nature of $\ \mathbb R$) we have
$$ f([x;y])\ \supseteq\ [f(x);f(y)] $$
Thus there exists $\ w\in[x;y]\ $ such that
$$ f(w)\ =\ \frac {f(x)+f(y)}2 $$
We see that $\ w\in(x;y)\ $ belongs to the interior of the interval, and $\ w\in f^{-1}[f(a);f(b)].\ $ This shows that $[a;b]\cap f^{-1}([f(a);f(b)])\ $ is dense in $\ [a;b],\ $ hence $\ [a;b]\subset f^{-1}(C).\ $ END of PROOF
After this exercise we get:
THEOREM 1 Let $\ f:\mathbb R^2\rightarrow \mathbb R\ $ be sliced. Then there exists $\ (a\ b)\in\mathbb R^2\setminus\{(0\ 0)\}\ $ such that:
$$\forall_{(x\ y)\,\ (x'\ y')\,\in\,\mathbb R^2 }\ \ \left(\ a\cdot x+b\cdot y = a\cdot x'+b\cdot y'\ \ \Rightarrow\ \ f(x\ y)=f(x'\ y')\ \right)$$
PROOF The case of a constant function is trivial. Otherwise there exists $\ h\in\mathbb R\ $ such that both sets $\ f^{-1}((-\infty\;h))\ $ and $\ f^{-1}((h;\infty))\ $ are non-empty. Then these two sets are disjoint open half-planes, i.e. they are non-empty, open and convex, and the complement of each of them is non-empty, closed and convex. Thus there exist $\ s\ t\in\mathbb R\ $ and $\ (a\ b)\in\mathbb R^2\setminus\{(0\ 0\}\ $ such that $\ s<t\ $ and
$$f(x\ y) = h\quad\Leftrightarrow\quad s\ \le\ a\cdot x+b\cdot y\ \le\ t$$
Then this $\ (a\ b)\ $ is the one required by the theorem. END of PROOF
This is about all about the general structure of the sliced functions.
A sliced function is constant in the direction perpendicular to the vector $\ (a\ b)\in\mathbb R\setminus\{(0\ 0)\}\ $ (see above), and it is monotone along that direction: let $\ \phi:\mathbb R\rightarrow\mathbb R\ $ be given by:
$$\forall_{u\in\mathbb R}\ \ \phi(u)\ :=\ f\left(u\cdot(a\ b)\right)$$
Then $\ \phi\ $ is monotonne. Thus each slide function is differentiable almost everywhere.
Re: Alex' comment on the OP, I assume "convex" w/r/t subsets of $\mathbb{R}^2$ is meant in the usual sense.
Call a function as in the OP tame. Here's an answer to (1):
First, I claim there is no total order $\prec$ on $\mathbb{R}^2$ such that any tame function is monotone with respect to $\prec$. To see this, consider the four points $a=(0, 1)$, $b=(1, 0)$, $c=(0, -1)$, and $d=(-1, 0)$. Then we must have:
Either every point on $l_1$ is $\prec$ every point on $l_2$ ("$l_1\prec l_2$"), or vice versa, and
Either every point on $l_3$ is $\prec$ every point on $l_4$ ("$l_3\prec l_4$"), or vice versa.
(To see this, consider the tame functions $f(x, y)=x+y$ and $g(x, y)=x-y$.) But now we reach a contradiction: suppose WLOG $l_1\prec l_2$ and $l_3\prec l_4$ - then $c\prec a$ since $l_1\prec l_2$, but $a\prec c$ since $l_3\prec l_4$.
As for partial orders, every function whatsoever is monotone w/r/t the discrete order; so maybe some constraint on the considered partial orders should be imposed?