How do we take second order of total differential?

$d=dx\frac{\partial}{\partial x}+dy\frac{\partial}{\partial y}$ is a differential operator that when applied to $f$ gives $df=dx\frac{\partial f}{\partial x}+dy\frac{\partial f}{\partial y}$

$d^2=\left(dx\frac{\partial}{\partial x}+dy\frac{\partial}{\partial y}\right)\left(dx\frac{\partial}{\partial x}+dy\frac{\partial}{\partial y}\right)$

$d^2f=\left(dx\frac{\partial}{\partial x}+dy\frac{\partial}{\partial y}\right)\left(dx\frac{\partial}{\partial x}+dy\frac{\partial}{\partial y}\right)f=\left(dx^2\frac{\partial^2}{\partial x^2}+dy^2\frac{\partial^2}{\partial y^2}+2dx dy\frac{\partial^2}{\partial x\partial y}\right)f=dx^2\frac{\partial^2 f}{\partial x^2}+dy^2\frac{\partial^2 f}{\partial y^2}+2dx dy\frac{\partial^2 f}{\partial x\partial y}$

I will assume that you are referring to the Fréchet derivative. If $U\subseteq\mathbb{R}^n$ is an open set and we have functions $\omega_{j_1,\dots,j_p}:U\to\mathbb{R}$, then $$ D\left(\sum_{j_1,\dots,j_p} \omega_{j_1,\dots,j_p} dx_{j_1}\otimes\cdots\otimes dx_{j_p}\right) = \sum_{j_1,\dots,j_p}\sum_{j=1}^n \frac{\partial\omega_{j_1,\dots,j_p}}{\partial x_{j}} dx_j\otimes dx_{j_1}\otimes\cdots\otimes dx_{j_p}. $$

Here $dx_{i}$ is the projection onto the $i$th coordinate, and if $\alpha,\beta$ are multilinear forms then $\alpha\otimes\beta$ is the multilinear form defined by $(\alpha\otimes\beta)(x,y)=\alpha(x)\beta(y)$.

For example, let $f(x,y)=x^3+x^2 y^2+y^3$. Then \begin{align} Df&=(3x^2+2xy^2)dx+(2x^2y+3y^2)dy; \\ D^2f&=(6x+2y^2)dx\otimes dx+4xy(dx\otimes dy+dy\otimes dx)+(2x^2+6y)dy\otimes dy; \\ D^3f&=6dx\otimes dx\otimes dx+4y(dx\otimes dx\otimes dy+dx\otimes dy\otimes dx+dy\otimes dx\otimes dx)\\ &\qquad+4x(dx\otimes dy\otimes dy+dy\otimes dx\otimes dy+dy\otimes dy\otimes dx) \\ &\qquad+6dy\otimes dy\otimes dy. \end{align}

Since $D^p f(x)$ is always a symmetric multilinear map if $f$ is of class $C^p$, you might want to simplify the above by using the symmetric product (of tensors).

Edit: I interpret your $d^2 f$ as $d^2f=d(df)$.

If $f$ is $C^2(\Omega)$, with $(x,y)\in\Omega$, then

$$d(df)=dy\wedge dx \frac{\partial^2 f}{\partial y\partial x}+dx\wedge dy \frac{\partial^2 f}{\partial x\partial y}=\left(\frac{\partial^2 f}{\partial x\partial y}- \frac{\partial^2 f}{\partial y\partial x} \right)dx\wedge dy=0,$$

as $dx \wedge dy=-dy\wedge dx$ and $\frac{\partial^2 f}{\partial x\partial y}= \frac{\partial^2 f}{\partial y\partial x}$ on $\Omega$.

If $f$ is not $C^2$ at $(x,y)$, then we arrive at

$$d(df)=\left(\frac{\partial^2 f}{\partial x\partial y}- \frac{\partial^2 f}{\partial y\partial x} \right)dx\wedge dy. $$