What is the exact meaning of the following sentence?

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 $\Phi'_-(x)<\Phi'_+(x)$.