Suppose $f(x)$ is convex, then we want to show that $f(x+tv), t\in \mathbb{R}, v \in \mathbb{R}^n, x \in \mathbb{R}^n$
To do this we can use the second derivative test:
(1) $\dfrac{d f(x+tv)}{dt} = f'(x+tv) v = \nabla f(x+tv)^T v$ (where we used $f'(x) = \nabla f(x)^T$)
(2) $\dfrac{d^2f(x+tv)}{dt^2} = (\nabla^2 f(x+tv)v)^Tv =v^T\nabla^2 f(x+tv)v $
It doesn't seem intuitive or obvious to me as to how the proof went from (1) to (2)
Can someone show me why is that $\dfrac{d\nabla f(x+tv)^T v}{dt} = (\nabla^2 f(x+tv)v)^Tv $
More specifically, how can we see that
$\dfrac{d\nabla f(x+t)}{dt} = \nabla^2 f(x+t)$
This answer assumes $f:\mathbb{R}^n\to \mathbb{R}$. For convenience I'll denote by $DF$ the total derivative or the Jacobian of $F$ (which in your question you denoted by "$F'$" which is just not convenient, but I'll try to include your notation as well). From your notation I assume that the gradient $\nabla$ is a column vector and $\nabla^2$ is the Hessian.
Prerequisites: You need to consider $C^2$-functions to use any derivative tests. So I'll assume that's the case.
The specific case: Note that you are differentiating w.r.t. $t$, so (after pushing the transpose out of the differential by "$\frac{d}{dt}(...^T)=(\frac{d}{dt} ...)^T$ to avoid confusion) careful use of chain rule yields $$\frac{d}{dt} (\nabla f(x+t)^T)=(\underbrace{(D(\nabla f))}_{=(\nabla f)'=\nabla^2f^T}\kern{-0.5em}(x+t)~\frac{d}{dt}(x+t))^T=(\nabla^2 f(x+t)^T~\mathbf{1})^T =~\mathbf{1}^T~\nabla^2f(x+t)$$ where $\mathbf{1}$ is the unit vector in $\mathbb{R}^n$. The first equality under the brace is just your notation for the total derivative of $f$. The second one can be derived from what you already used in your question in (1) ("where we used [...]" with $\nabla f$ instead of $f$) or from Wikipedia ("The Hessian matrix can be considered related to the Jacobian matrix by [...]" with $\nabla^2$ instead of $H$).
The application: Using the same principle yields for fixed $x,v\in\mathbb{R}^n$: $$\begin{align}\dfrac{d\nabla f(x+tv)^T v}{dt} &= \frac{d}{dt}\left(\nabla f(x+tv)\right)^Tv\\ &=\big(\underbrace{(D(\nabla f))}_{=\nabla^2f^T~\text{as above}}\kern{-0.6em}(x+tv)~ \underbrace{\frac{d}{dt}(x+tv)}_{=v}\big)^T v\\ &=(\nabla^2 f(x+tv)v)^Tv\end{align}$$
Conclusion: Suppose $f$ is convex. This is equivalent to the Hessian $\nabla^2f$ being positive definite which means $$\frac{d^2 f(x+tv)}{dt^2}=v^T\nabla^2f(x+tv)v\geq 0.$$ This means that $f$ is convex along the line $\{x+tv\}_{t\in\mathbb{R}}$.
(I tidied up and tried to adapt more to your notation. If something's unclear or if you feel I misunderstood parts of your question, please feel free to point that out!)