Can "being differentiable" imply "having continuous partial derivatives"?

No, there are differentiable functions which are not $C^1.$ For example consider $f:\mathbb{R} \rightarrow \mathbb{R}$ given by $f(x) = x^2 \sin(\frac{1}{x})$ for $x \neq 0$ and $f(x) = 0$ if $x = 0.$ Then $f$ is differentiable but not $C^1.$

To see that $f$ is differentiable away from zero is not hard and follows from the product and chain rules. On the other hand to show $f$ is diffentiable at $0$ we must appeal to the definition of the derivative.

Observe

$$\displaystyle\lim_{x\rightarrow 0} \frac{f(x) - f(0)}{x-0} = \displaystyle\lim_{x\rightarrow 0} \text{ } x \sin(1/x)$$

which converges to $0$ as $|x \sin(1/x)| < |x|$.

Hence, $f$ is everywhere diffentiable.

It follows $f'(x)$ is given by $2x\sin(\frac{1}{x}) - \cos(\frac{1}{x})$ if $x \neq 0$ and $0$ if $x = 0.$

We claim $f'$ is not continuous at $0.$ To see this, note that if $f'$ were continuous at $0,$ then so too would be the function $g(x)$ defined by $\cos(1/x)$ if $x \neq 0$ and $0$ if $x = 0.$ But this is false as $\langle\frac{1}{\pi + n2\pi}\rangle_{n\in\mathbb{Z}_+}$ is a sequence of real numbers converging to $0$ such that

$$\displaystyle\lim_{n\rightarrow \infty} g\left(\frac{1}{\pi + n 2\pi}\right) = 1 \neq 0 = g(0).$$

Hence, $f$ is differentiable but not $C^1.$

Re your first question, the converse is false and one most usually mentions counterexamples based on the oscillatory behaviour of $x\mapsto\sin(1/x)$ near $x=0$.

Re your second question, Chern's and Folland's definitions are identical since both require $D^kf$ to be continuous for $f$ to belong to $C^k$, as they should.

Edit Due to a radical modification of the question, the above only partly applies to the current version of the question. Great...

However, one can say this: the case $m\ge2$ cannot yield stronger results, simply consider a function like $(f,0,0,\ldots,0)$ where $f$ is a counterexample when $m=1$.