The definition simply says that the dynamics of a system does not change after a symmetry transformation. By "equal in form" he means that if I apply the transformation and then I rename the new variables as the old ones, the new and the old equations are identical.
The two parts of the definition are equivalent because:
A symmetry sends a solution to another solution $\Longrightarrow$ A symmetry sends an equation of motion (EOM) to an equation that has the same solutions, i.e. the same equation.
A symmetry sends an equation to the same equation $\Longrightarrow$ A symmetry sends a solution of the old EOM to a solution of the new EOM. But, in fact, the two EOMs are the same.
Example of a symmetry
An isolated system has a Lagrangian
$$
\mathcal{L} = \sum_{i=1}^n \frac12 {\dot{\vec{q}}_i}^2 + V(|\vec{q}_k-\vec{q}_j|^2)\,.
$$
Therefore the equations of motion will be of the form (recall $\partial_{\vec{q}_i} |q_i-q_j|^2 = 2 (\vec{q}_i - \vec{q}_j)$)
$$
\ddot{\vec{q}}_i = \sum_j(\vec{q}_i -\vec{q}_j)\,W_j(|\vec{q}_k-\vec{q}_j|^2)\,.
$$
If I make a rotation $\vec{q}_i \to R\cdot\vec{q}_i$ such that $R^TR=1$ then the arguments of $W$ remain all unchanged because it's made of scalar products and the EOM reads
$$
R\cdot\ddot{\vec{q}}_i = \sum_j(R\cdot\vec{q}_i - R\cdot\vec{q}_j )\,W_j(|\vec{q}_k-\vec{q}_j|^2)\,.
$$
Now if I just call $R\cdot\vec{q} = \vec{q}$ again, the equation is the same. The solutions can be as complicated as you want by choosing $V$ to be an ugly function. However, if I make a rotation that's the same as measuring all coordinats with a rotated frame. And since the system is isolated the frame is arbitrary anyway. Therefore if me and a friend of mine measure things with rotated frames and we look at a physical trajectory of a particle, both of us better conclude that the trajectory is a solution to the EOM.
Example of something that is not a symmetry
Just take the example above and add an explicit vector $\vec{X}$ coupled to one of the $\vec{q}$'s, like
$$
\mathcal{L} \to \mathcal{L} + \vec{X} \cdot \vec{q}_1\,.
$$
If you make a rotation, the old equations will look like the new ones, except for the fact that the new ones have the vector $R^T\cdot \vec{X}$ rather than $\vec{X}$. Accordingly, a solution of the original system won't be a solution of the rotated one.
And why is it physically acceptable that here rotations are not a symmetry? Because we had to choose a specific vector $\vec{X}$. So now that makes the choice of frame not arbitrary anymore: there is a frame that is preferred with respect to the other ones and it's the one where $\vec{X}=(1,0,0)$. Or, to put it in other terms, I can tell apart two configurations that differ by a rotation by checking at the direction of $\vec{X}$ in the two frames. This breaks the symmetry.
