I have read the following sentence in my book.
Let $\Phi$ be a convex function which is finite on interval $I$ containing the range of the function $g$ and infinite elsewhere, and let $\phi$ be an arbitrary determination of its derivative (i.e. having at a "corner" any value between the left and right derivatives), defined and finite on $I$.
What is the meaning of "having at a 'corner' any value between the left and right derivatives"?
The only way a convex function can fail to be differentiable at a point is to have a "corner" like the one for $|x|$ at $0$. The derivative of that function is $-1$ for $x < 0$ and $+1$ for $x > 0$. Those are the values of the left and right derivatives at $0$. Since they are different there is no derivative at $0$. The quotation says that for the purposes it has in mind you can choose any value between $-1$ and $+1$ for the derivative at $0$.
A "corner" occurs if the derivate from the left differs from the derivate from the right.
At such positions, the function is not differentiable.
A simple example is $f(x)=|x|$ for $x_0=0$.
For $x>0$ , we have $f(x)=x$, hence $f'(x)=1$
For $x<0$ , we have $f(x)=-x$, hence $f'(x)=-1$
So, the limit for $x\rightarrow 0$ does not exist.
The convex function $\Phi$ has right and left derivatives $\Phi'_+(x)$ and $\Phi'_-(x)$ at each point $x$ in the interior of $I$. Both $\Phi'_+$ and $\Phi'_-$ are non-decreasing, $\Phi'_+(x-)=\Phi'_-(x)$ and $\Phi'_-(x+)=\Phi'_+(x)$ for all $x$ in the interior of $I$. It follows that $\Phi'_-(x)\le\Phi'_+(x)$ with equality except on the countable set of $x$ values where $\Phi'_+$ has a jump discontinuity. A "corner" refers to an $x$ with $\Phi'_-(x)<\Phi'_+(x)$. It seems that the author of your book would have you take $\phi(x)=\Phi'_+(x)$ if $\Phi'_+(x)=\Phi'_-(x)$, and $\phi(x)$ any point of $[\Phi'_-(x),\Phi'_+(x)]$ if $\Phi'_-(x)<\Phi'_+(x)$.
The statement that $\phi$ is "an arbitrary determination of its derivative" means that if $x \in I$, then $\phi(x) \in \partial \Phi(x)$, where $\partial \Phi(x)$ is the subdifferential of $\Phi$ at $x$.
Edit:
I should have explained what "subdifferential" means. If a convex function $f:\mathbb R^n \to \mathbb R \cup \{\infty\}$ is finite and differentiable at $x_0$, then it can be shown that $f(x) \geq f(x_0) + \langle \nabla f(x_0),x - x_0 \rangle$ for all $x \in \mathbb R^n$. Even if $f$ is not differentiable at $x_0$, usually there exist vectors $g$ such that $$ \tag{$\spadesuit$} f(x) \geq f(x_0) + \langle g, x - x_0 \rangle \quad \text{for all } x \in \mathbb R^n. $$ A vector $g$ which has this property is called a "subgradient" of $f$ at $x_0$. So even if $f$ does not have a gradient at $x_0$, at least $f$ has subgradients at $x_0$, and that is better than nothing.
The set of all subgradients of $f$ at $x_0$ is called the subdifferential of $f$ at $x_0$, and is denoted $\partial f(x_0)$. For example, let $f:\mathbb R \to \mathbb R$ such that $f(x) = | x |$ for all $x \in \mathbb R$. Then $\partial f(0) = [-1,1]$.
It can be shown that, under mild assumptions, a convex function $f$ is differentiable at $x_0$ if and only if $\partial f(x_0)$ is a singleton. In this case, $\partial f(x_0) = \{ \nabla f(x_0) \}$.
The right and left derivatives of $\Phi$ at a point $a$ in the domain of $\Phi$ are respectively \begin{align} & \lim_{\Delta x\,\downarrow\,a} \frac{ \Phi(a+\Delta x) - \Phi(a) }{\Delta x} \\[10pt] \text{and } & \lim_{\Delta x\,\uparrow\,a} \frac{ \Phi(a+\Delta x) - \Phi(a) }{\Delta x} \end{align} With a convex function, the left derivative at a point is always less than or equal to the right derivative. At most points they are equal. A point where they are not equal is a "corner". If there is a corner at which the left derivative is $3$ and the right derivative is $4$, then they're assigning to $\varphi$ at that point some arbitrary value not less than $3$ and not greater than $4$.