The idea is to consider a suitable $\mathbb{R}[Y]$-linear combination of $f_0(X)$ and $f_2(1 - X)$, namely, $(1 - Y)f_0(X) + Yf_2(1 - X)$, and a suitable $\mathbb{R}[X]$-linear combination of $f_1(Y)$ and $f_3(1-Y)$, namely, $Xf_1(Y) + (1-X)f_3(1-Y)$, along with a suitable degree $2$ corrective term, $a_{11}XY + a_{10}X + a_{01}Y + a_{00}$, collect them together to form
$$f(X, Y) = (1-Y)f_0(X) + Xf_1(Y) + Yf_2(1-X) + (1-X)f_3(1-Y) + a_{11}XY + a_{10}X + a_{01}Y + a_{00}$$
and require the latter to satisfy the conditions in the statement.

Thus, the condition $f(X, 0) = f_0(X)$ requires $a_{10} = f_3(1) - f_1(0)$ and $a_{00} = -f_3(1)$. Refer to the hypothesis on the $f_k$ to write $a_{10} = f_0(0) - f_1(0)$ and $a_{00} = -f_0(0)$.
Next, the condition $f(0, 1-Y) = f_3(Y)$ requires $a_{01} = f_0(0) - f_2(1)$ and $f_2(1) + a_{01} + a_{00} = 0$. Notice that the latter holds automatically. With reference again to the hypothesis on the $f_k$, $a_{01} = f_0(0) - f_3(0)$.
Similarly, the condition $f(1, Y) = f_1(Y)$ requires $a_{11} = f_0(1) - f_2(0) - a_{01}$ and $f_0(1) + a_{10} + a_{00} = 0$. The latter holds again automatically, while the former yields $a_{11} = f_1(0) - f_2(0) - f_0(0) + f_3(0)$.
Finally, the condition $f(1-X, 1) = f_2(X)$ requires $f_3(0) - f_1(1) - a_{11} - a_{10} = 0$ and $f_1(1) + a_{11} + a_{10} + a_{01} + a_{00} = 0$, both of which hold by the preceding and the hypothesis on the $f_k$.
In terms of the data, the desired polynomial is
$$f(X, Y) = (1 - Y)f_0(X) + Xf_1(Y) + Yf_2(1 - X) + (1 - X)f_3(1 - Y) - (f_0(0) - f_1(0) + f_2(0) - f_3(0))XY + (f_0(0) - f_1(0))X + (f_0(0) - f_3(0))Y - f_0(0).$$

The polynomial in the previous solution may equally well be obtained as follows: begin by seeking a two-variable polynomial $f$ of the form
$$f(X, Y) = g(X, Y) + Xh(Y) + Yk(X) + aXY,$$
where $g$ is a two-variable polynomial satisfying $g(X, 0) = f_0(X)$ and $g(0, 1-Y) = f_3(Y)$, $h$ and $k$ are one-variable polynomials subject to $h(0) = k(0) = 0$ and $h(1) = k(1)$, and $a$ is a real number to be determined from the requirements. The common value $h(1) = k(1)$ will come out of the choice of $h$ and $k$.
Notice that such a choice of $g$, $h$ and $k$ yields $f(X, 0) = f_0(X)$ and $f(0, 1-Y) = f_3(Y)$.
Now, the two-variable polynomial $g(X, Y) = f_0(X) + f_3(1-Y) - f_0(0)$ clearly fits the bill; with reference to the hypothesis on the $f_k$, the verification is routine and hence omitted.
Next, letting $h(Y) = f_1(Y) - g(1, Y)$ and $k(X) = f_2(1-X) - g(X, 1)$, it is again readily checked that $h(0) = k(0) = 0$, and $h(1) = f_1(1) - g(1, 1) = f_2(0) - g(1, 1) = k(1)$. In terms of the data, $h(Y) = f_1(Y) - f_3(1-Y) + f_0(0) - f_1(0)$, $k(X) = -f_0(X) + f_2(1-X) + f_0(0) - f_3(0)$, and $h(1) = k(1) = f_0(0) - f_1(0) + f_2(0) - f_3(0)$.
At this stage, requiring $f(1, Y) = f_1(Y)$ yields $a = -f_0(0) + f_1(0) - f_2(0) + f_3(0)$; incidentally, yet not accidentally at all, this is precisely the value of $a_{11}$ in the previous solution.
Finally, notice that $h(1) = -a$, to check the remaining requirement: $f(1-X, 1) = g(1-X, 1) + h(1)(1-X) + k(1-X) + a(1-X) = g(1-X, 1) + k(1-X) = f_2(X)$.

Expressing $f$ in terms of the data yields the polynomial in the previous solution, just as mentioned in the beginning.