1. In equation ii/ we see that
$$(x-1)-y \equiv (x+1)-y \equiv x-(y-1) \equiv x-(y+1) \pmod{2}$$
That is, the values of $f(x,y)$ where $x,y$ have the same parity and the values of $f(x,y)$ where $x,y$ have the opposite parity have no connection.
We will find the way to define function $f$ in both cases.
In the coordinate plane $Oxy$, we consider the lattice points $(i, j)$ where $i, j$ are integers and $0 \le j \le 2n+1$. The point $(i, j)$ is assigned with the value $f(i, j)$.
The condition i/ gives us that all numbers in the border (upper and lower) are 0; and all inner points are assigned with the numbers from $[0; 1]$.

The condition ii/ means for each small skew square (with diagonal parallel to the axis), with vertices in the lattice points and with side $\sqrt{2}$, the sum of all assigned numbers is equal to 1.
For each lattice point $A(i, j)$, let $f_1(A) = f(i+2,j)$, $f_2(A) = f(i-2,j)$.
Let $a_k = f(k, k)$ where $k=1,2,3,\ldots,2n$ be the value assigned to $A_k(k, k)$. By the condition, we have
$$\begin{align*} a_1 + a_2 + f_1(A_1) + 0 &= 1 & \Rightarrow f_1(A_1) &= 1 - a_1 - a_2, \\ a_2 + a_3 + f_1(A_2) + f_1(A_1) &= 1 & \Rightarrow f_1(A_2) &= a_1 - a_3, \\ \dots & & \end{align*}$$
$$\begin{align*} a_{2n-1} + a_{2n} + f_1(A_{2n-2}) + f_1(A_{2n-1}) &= 1 & \Rightarrow f_1(A_{2n-1}) &= a_1 - a_{2n}, \\ a_{2n} + 0 + f_1(A_{2n-1}) + f_1(A_{2n}) &= 1 & \Rightarrow f_1(A_{2n}) &= a_1 \end{align*}$$
Similarly, we have
$$\begin{align*} f_2(A_{2n}) &= 1 - a_{2n-1} - a_{2n} \\ f_2(A_{2n-1}) &= a_{2n} - a_{2n-2} \\ \dots & \end{align*}$$
$$\begin{align*} f_2(A_1) &= 1 - a_1 - a_{2n}, \\ f_2(A_2) &= a_{2n} \end{align*}$$
From that we see that if the sequence $a_k$ is given then the values of $f_1(A_k), f_2(A_k)$ are defined. We will choose $a_k$ such that $f_1(A_k), f_2(A_k)$ belong to $[0;1]$.
We choose $a_1 \ge a_3 \ge a_5 \ge \dots \ge a_{2n-1}, a_2 \le a_4 \le a_6 \le \dots \le a_{2n}, a_1 + a_{2n} \le 1$ then $f_1(A_k), f_2(A_k) \in [0;1]$.
Moreover, we see that
$$\begin{align*} f_1(A_1) &\ge f_1(A_3) \ge \dots \ge f_1(A_{2n-1}), f_1(A_2) \le f_1(A_4) \le \dots \le f_1(A_{2n}), f_1(A_1) + f_1(A_{2n}) \le 1, \\ f_2(A_1) &\ge f_2(A_3) \ge \dots \ge f_2(A_{2n-1}), f_2(A_2) \le f_2(A_4) \le \dots \le f_2(A_{2n}), f_2(A_1) + f_2(A_{2n}) \le 1 \end{align*}$$
So, the monotonicity of two subsequences of odd indices and even indices still holds for $f_1(A_k)$ and $f_2(A_k)$, as well as the last condition.
Thus, similarly, from the sequence $f_1(A_k)$, we can find $f_1(f_1(A_k))$, and from $f_2(A_k)$, we can find $f_2(f_2(A_k))$ and so on, and we can find all values of $f(x,y)$ where $x-y$ is even.

By denoting $b_k = f(k+1, k)$ the values assigned to $B_k(k+1, k)$, where $k=1,2,3,\ldots,2n$ then by similar way as below, we can construct all values of $f(x,y)$ where $x-y$ is odd.
It is easy to see that there are infinitely many ways to choose sequences $(a_k), (b_k)$ which satisfy the following condition. Thus there exists infinitely many functions with the given properties, i.e. $|F|$ is infinite. (Q.E.D)

2. In the equation $f(x-1,y)+f(x+1,y)+f(x,y-1)+f(x,y+1)=1$, replacing $x,y$ by $x+1,y+1$, we get $f(x,y+1)+f(x+2,y+1)+f(x+1,y)+f(x+1,y+2)=1$.
Therefore $f(x-1,y)+f(x,y-1) = f(x+1,y+2)+f(x+2,y+1)$ (*) or
$$f(1,1)+f(2,0) = f(3,3)+f(4,2) = \dots = f(2n+1,2n+1)+f(2n+2,2n),$$
$$f(3,1)+f(4,0) = f(5,3)+f(6,2) = \dots = f(2n+3,2n+1)+f(2n+4,2n).$$
Thus $f(1,1) = f(2n+2,2n)$, $f(3,1) = f(2n+4,2n)$.
Similarly, we have $f(2,2) = f(2n+2,2n-1)$, $f(4,2) = f(2n+5,2n-1)$.
Repeatedly apply (*), we have
$$f(k,k) = f(2n+1+k, 2n+1-k), f(k+2,k) = f(2n+3+k, 2n+1-k).$$
Similarly, we have
$$f(2n+k, 2n+1-k) = f(4n+2+k, k), f(2n+3+k, 2n+1-k) = f(4n+4+k, k).$$
Thus, $f(k,k) = f(2n+1+k, 2n+1-k) = f(4n+2+k, k)$ for all $k=1,2,3,\ldots,2n$. And therefore, by induction, we get
$$f(k,k) = f((2n+1)i+k, 2n+1-k) = f((4n+2)i+k, k) \text{ where } i \in \mathbb{Z} \text{ and } k=1,2,3,\ldots,2n.$$
Moreover, the definition of values in the next diagonal (to the right) is the same in the diagonals $(k,k)$ and $((2n+1)i+k, 2n+1-k)$ so the values at respective points are equal. Therefore values of $f$ are repeated as illustrated in the picture, i.e. the values of $f(x,y)$ with $x-y$ even repeat the values assigned to the lattice points inside triangle $\Omega$ with vertices at $(1,1), (2n+1,2n), (4n+1,1)$.

Number of such values is finite and we can compute that there are at most
$$1+2+3+\ldots+2n = n(2n+1) \text{ distinct values.}$$
Similarly with the values of $f(x,y)$ where $x-y$ is odd, in this case we also have at most $n(2n+1)$ distinct values. Besides that, we have 0 as one value of $f(x, y)$. Summing all above arguments, we have $|v_f| \le 2n(2n+1)+1$.
Thus, $|v_f|$ is finite for all $f \in F$, and we have done.

3. We will construct a function $f$ for which $|v_f| = 2n(2n+1)+1$ and show that this is the maximum value of $|v_f|$.
By induction, we can prove that
$$f(i+2k,i) = \frac{(1-(-1)^i)(1-(-1)^k)}{4} + (-1)^k a_{k+i} - (-1)^{k+i} a_k$$
for all $i=1,2,\ldots,2n$ and $k \in \mathbb{Z}$ such that $0 \le k+i \le 2n+1$. (**)
Indeed, for $k=0$ then () is obviously true.
Assume that () is true for all $(i,k)$ with $k \le m$ and $k=m+1, i \le j$. From the condition ii/, we have
$$f(j+1+2(m+1), j+1)+f(j+1+2m, j+1)+f(j+2(m+1), j)+f(j+2+2m, j+2)=1.$$
By the induction hypothesis, we can compute
$$f(j+1+2(m+1), j+1) = \frac{(1-(-1)^{j+1})(1-(-1)^{m+1})}{4} + (-1)^{m+1}a_{m+j+2} - (-1)^{m+j+2}a_{m+1}.$$
Therefore (**) is true for $i = j+1, k = m+1$.
Thus, the values assigned to the points inside the triangle $\Omega$ have the form $\delta_{ij} \pm a_i \pm a_j$ with $\delta_{ij} \in \{0,1\}$, where $\delta_{ij}$ as well as the signs of $a_i, a_j$ are well defined by $i, j$.
Next, we choose $a_{2k-1} = \frac{1}{3^{2k-1}}, a_{2k} = \frac{1}{3^{2(n+1-k)}}$ where $k=1,2,\ldots,n$ then using the fact that every integer can be expressed in the form $\sum_{i=0}^{r} \delta_i 3^i$ where $\delta_i \ne 0$ and $e_i \in \{-1,0,1\}$ in only one way, we see that all values $f(x,y)$ assigned to the points inside triangle $\Omega$ are distinct and are non-zero, i.e. we have exactly $n(2n+1)$ such values. We have constructed the values of $f(x,y)$ where $x-y$ is even.
Similarly, for construction of $f(x,y)$ where $x-y$ is odd, we choose
$$b_{2k-1} = \frac{1}{3^{2k-1}\sqrt{3}}, b_{2k} = \frac{1}{3^{2(n+1-k)}\sqrt{3}} \text{ where } k=1,2,\ldots,2n.$$
then the values of the form $\delta'_{ij} \pm b_i \pm b_j$ are also non-zero, distinct and different from the values $\delta_{ij} \pm a_i \pm a_j$.
From which, we construct all values for $f(x,y)$ and we have at all $2n(2n+1)+1$ distinct values.
Thus, the maximum value of $|v_f|$ on $F$ is $2n(2n+1)+1$.