Notation of the Taylor Polynomial with Lagrange Remainder I have this Theorem in my book: 

Consider $f: \mathbb{R^n} \rightarrow \mathbb{R}$ a function of class $C^1$ and $\overline{x}, d \in \mathbb{R^n}$. If $f$ is twice differentiable in the segment $(\overline{x}, \overline{x}+d)$, then exist $t \in (0,1)$ such that
$$
f(\overline{x}+d) = f(\overline{x}) + \nabla f(\overline{x})^{T}d + \dfrac{1}{2}d{^T} \nabla^2f(\overline{x}+td)d. 
$$

So, I couldn't understand the last term notation. I don't know because this is valid for some $t \in (0,1)$ and I don't understand the notation.
For example, I know that if I consider in $\mathbb{R^2}$ the pair $(x,y) = \overline{x} + d$ and $\overline{x} = (a,b)$, I have
$$\begin{align}
f(x,y) &= f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b) + \\ &+\dfrac{1}{2}[f_{xx}(a,b)(x-a)^2 + 2f_{xy}(a,b)(x-a)(y-b) + f_{yy}(y-b)^2].
\end{align}$$
And, the second term I can write 
$$
f_x(a,b)(x-a) + f_y(a,b)(y-b) = \nabla f(a,b)^T
\begin{bmatrix} x-a \\y -b \end{bmatrix} = 
\nabla f(a,b)^T
\left( \begin{bmatrix} x \\y  \end{bmatrix} - \begin{bmatrix} a \\b   \end{bmatrix} \right).
$$
So, If I get $x = \overline{x}+d$ I have 
$$
f(\overline{x} + d) = f(\overline{x}) + \nabla f(\overline{x})^T(x-\overline{x}) = f(\overline{x}) + \nabla f(\overline{x})^Td. 
$$
Now I can't understand the last term notation.
Could someone please explain the last notation and about the choose with $t$ in the last term of theorem in human language?

What you wrote doesn't make sense. I hope your book isn't writing the last term like that, or the author is using a strange notation. 
What you should have is something like this $${1\over 2} d^T H_f(\overline{x}) d$$ where $H_f(\overline{x})$ is the Hessian matrix which is the matrix of all the second order partial derivatives of a function. For a function $$f:\mathbb{R}^2\rightarrow\mathbb{R}$$ the Hessian is of the form $$H_f(\overline{x}) = \begin{bmatrix}\frac{\partial^2 f}{\partial x^2}&\frac{\partial^2 f}{\partial x\partial y} \\ \frac{\partial^2 f}{\partial y\partial x}& \frac{\partial^2 f}{\partial y^2}\end{bmatrix}$$ Most of the times, for Schwartz's theorem the off diagonal element are the same. Now you can see how this term make's sense. 
The notation used by the book is, most of the times, the Laplace's operator which is a totally different thing.
