Answer: 556

Small Hint:

The rightward jumps are $+1$s and $+2$s summing to $4,$ so as a multiset they are $\{1,1,1,1\},$ $\{1,1,2\},$ or $\{2,2\},$ and likewise for the upward jumps

Big Hint:

For each pair of multisets, the number of ways to order all the jumps is a multinomial coefficient; add up the nine cases

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 $$\begin{aligned} &\binom{8}{4} = 70, \\ &\quad \frac{7!}{4!\,2!} = 105, \\ &\quad \frac{6!}{4!\,2!} = 15, \end{aligned}$$ $$\begin{aligned} &\frac{6!}{2!\,2!} = 180, \\ &\quad \frac{5!}{2!\,2!} = 30, \\ &\quad \binom{4}{2} = 6. \end{aligned}$$ The values $105,$ $15,$ and $30$ each occur twice.

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