Answer: 556

Solution:

The horizontal jumps are steps of $1$ or $2$ summing to $4,$ so as a multiset they are $\{1,1,1,1\},$ $\{1,1,2\},$ or $\{2,2\},$ and the same holds for the vertical jumps. For any choice of the two multisets, every ordering of all the jumps is a valid sequence, and the number of orderings is the multinomial coefficient of the combined multiset.

The nine cases give $$\begin{aligned} &\binom{8}{4} = 70, \\ &\quad \frac{7!}{4!\,2!} = 105 \text{ (twice)}, \\ &\quad \frac{6!}{4!\,2!} = 15 \text{ (twice)}, \end{aligned}$$ $$\begin{aligned} &\frac{6!}{2!\,2!} = 180, \\ &\quad \frac{5!}{2!\,2!} = 30 \text{ (twice)}, \\ &\quad \binom{4}{2} = 6. \end{aligned}$$

The total is $70 + 2 \cdot 105 + 2 \cdot 15$ $+ 180 + 2 \cdot 30 + 6$ $= 556.$