Student claims $\lim_{x \to 0^+} \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = \lim_{h \to 0} \lim_{x \to 0^+} \frac{f(x+h)-f(x)}{h}$. Is it justified?

The second "$=$" in the students' answers is an invalid assumption. Interchanging the order (e.g of $h$ and $x$ ) in the limits often fails to preserve values. To justify a change of order requires additional work.

Suppose we were not given that $\lim_{x\to 0}f'(x)$ exists. Read the students' answer in reverse order. The reversal of the order of limits would then seem to imply that $f'$ is continuous at $0$ merely from $f$ being differentiable, which is a false conclusion.

So ask them: How & where did you use the existence of $\lim_{x\to 0^+}f'(x)$ to justify the reversal? The correct answer is "Nowhere".

Blindly interchanging the limits without any justifications cannot not be accepted as a valid answer. Here is a valid proof: For any $x>0$ there exits $\xi_x \in (0,x)$ such that $\frac {f(x)-f(0)} x =f'(\xi_x)$. If $\epsilon >0$ and $\delta$ is chosen such that $|f'(y)-L| <\epsilon$ for $0 <y <\delta$ then we get $|\frac {f(x)-f(0)} x-L| <\epsilon$ whenever $0 <x <\delta$ and this proves that $f'(0)=L$.

@glowstonetrees say in their answer

Perhaps it's true that $$ \lim_{x \to 0^+} \lim_{h \to 0} g(x,h) = \lim_{h \to 0} \lim_{x \to 0^+} g(x,h) $$ if $g(x,h)$ takes the form $g(x,h) = \frac{f(x+h) - f(x)}{h}$, . . .

I will show that even this need not be the case in general. Let $f \colon [0,1] \to \mathbb{R}$ be the function defined by $$ f(x) = \begin{cases} 0, &x = 0;\\ x^2 \sin(1/x), &x \neq 0. \end{cases} $$ Then, on the one hand we have $$ \lim_{h \to 0^+} \lim_{x \to 0^+} \frac{f(x+h)-f(x)}{h} = \lim_{h \to 0^+} \frac{f(h)-f(0)}{h} =f'(0) = 0. $$ On the other hand, we have $$ \lim_{x \to 0^+} \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{x \to 0^+} f'(x) = \lim_{x \to 0^+} 2x \sin(1/x) - \cos(1/x), $$ which does not exist.

So, the interchange of the two limits cannot be done simply by knowing that $g(x,h)$ is of the form $\frac{f(x+h) - f(x)}{h}$.