We suppose that the square is placed into a Cartesian system of coordinates with origin $A(0,0)$ and axes on the sides $AB$ and $AD$, as in the figure 7. Let the spider start from $A$ and make its first movement ($m$ steps right and $k$ steps up or $m$ steps up and $k$ right). In the figure you can see the case $m = 4, k = 3$.

After the first movement the spider can be on the point $M(k,m)$ or on the point $N(m,k)$. The number of ways the spider can approach the point $M(k,m)$ or the point $N(m,k)$ are equal to the combinations of $k+m$ elements taken $k$ at a time, that is
$$\binom{k+m}{k} = \binom{k+m}{m} = v.$$

After the second movement the spider can be at three different points: $K(2k, 2m)$, $L(k+m, k+m)$, and $T(2m, 2k)$ with corresponding number of ways $v^2$, $2v^2$, and $v^2$.
The point $M$ can be approached by $v$ different ways from $A$. The point $K$ can be approached by $v$ different ways from $M$. Hence the point $K$ can be approached by $v^2$ different ways from point $A$.

In a similar fashion we can find that point $T$ can be approached by $v^2$ different ways from point $A$. The point $L$ can be approached by $v^2$ different ways by following the route $A, M, L$ and by $v^2$ different ways by following the route $A, N, L$. Hence point $L$ can be approached by $2v^2$ different ways from point $A$.

Similarly, we conclude that after the third movement the spider can be placed in four different points that can be approached by $\binom{3}{3} \cdot v^3$, $\binom{3}{2} \cdot v^3$, $\binom{3}{1} \cdot v^3$, and $\binom{3}{0} \cdot v^3$ different ways, respectively. Therefore, after the completion of the $l$ movements, the spider will be placed at $l+1$ different points, lying on the line with equation: $x + y = l(k+m)$.

If $r = m-k$, then these points are the following:
$$A_0(lm, lk),\ A_1(lm-r, lk+r),\ A_2(lm-2r, lk+2r),\ \dots,\ A_l(lm-lr, lk+lr) \equiv A_{l}(lk, lm).$$
These points can be approached by $\binom{l}{l} \cdot v^l$, $\binom{l}{l-1} \cdot v^l$, $\binom{l}{l-2} \cdot v^l$, $\dots$, and $\binom{l}{0} \cdot v^l$ different ways from the point $A$. The point $C(ml, ml)$ can be approached from the point:

$$A_0(lm, lk),\ \text{ with } \binom{(ml-lm)+(ml-lk)}{ml-lk} = \binom{ml-lk}{ml-lk} = \binom{lr}{lr} = 1, \text{ way.}$$
$$A_1(lm-r, lk+r),\ \text{ with } \binom{(ml-lm+r)+(ml-lk-r)}{r} = \binom{ml-lk}{r}, \text{ ways.}$$
$$A_2(lm-2r, lk+2r),\ \text{ with } \binom{(ml-lm+2r)+(ml-lk-2r)}{2r} = \binom{ml-lk}{2r}, \text{ ways.}$$
$$A_l(lk, lm),\ \text{ with } \binom{(ml-lk)+(ml-lm)}{ml-lk} = \binom{lr}{lr} = 1, \text{ way.}$$
Hence the point $C(ml, lm)$ can be approached from point $A$ by:
$$v^l \left( \binom{l}{l} \cdot \binom{lr}{0} + \binom{l}{l-1} \cdot \binom{lr}{r} + \binom{l}{l-2} \cdot \binom{lr}{2r} + \cdots + \binom{l}{0} \cdot \binom{lr}{lr} \right) \text{ ways.}$$

(*) Moving from the point $(i, j)$ to the point $(i+k, j+m)$ of the grid with direction right and up, can be made by $\binom{k+m}{k} = \binom{k+m}{m}$ different ways. The procedure of finding the number of ways is the same as the procedure we have used at the first movement.

Note. In the case where $m-k = r = 1$, then the number of all possible ways becomes:
$$v^l \left( \binom{l}{l} \cdot \binom{l}{0} + \binom{l}{l-1} \cdot \binom{l}{1} + \binom{l}{l-2} \cdot \binom{l}{2} + \cdots + \binom{l}{0} \cdot \binom{l}{l} \right) = v^l \binom{2l}{l} = \binom{k+m}{m}^l \binom{2l}{l}.$$