My questin is as the title, why the Euler characteristic is additive on a short exact sequence, that is,
If we have a triple $(X,Y,Z)$ for $Z\subset Y\subset X$,then we have $$ \chi(X,Z)=\chi(X,Y)+\chi(Y,Z) $$ where the Euler characteristic is $$ \chi(X,Y)=\sum_{i=1}^{n}(-1)^i rank\,H_i(X,Y) $$ for a pair $(X,Y)$.
I do not figure out the more details!
Please help me,thank you！
Here is a linear algebra exercise, which I leave to you to prove. Whenever you have a short exact sequence of vector spaces, $$0 \to V_1 \to V_2 \to V_3 \to 0,$$ then one finds that the alternating sum of dimensions is zero: $\dim V_2 - \dim V_1 - \dim V_3 = 0$. (That is, the middle term is the sum of the dimensions of the outer two.)
This generalizes in fact to much longer sequences: if $$0 \to V_1 \to V_2 \to \cdots \to V_n \to 0$$ is an exact sequence of vector spaces, one has $$\dim V_1 - \dim V_2 + \dim V_3 - \cdots + (-1)^{n+1} \dim V_n = 0.$$
What you want to apply this to is the long exact sequence of a triple, $$\cdots \to H_{i+1}(X,Y) \to H_i(Y, Z) \to H_i(X, Z) \to H_i(X, Y) \to H_{i-1}(X,Y) \to \cdots$$
You have assumed in your question that this sequence stops eventually (that is, $H_{n+1}(X,Y) = H_{n+1}(X,Z) = H_{n+1}(Y,Z) = 0$.) This means we can apply the linear algebra statement above.
As a notational point, from here on I will write $\dim H_k(X,Y) = b_k(X,Y)$ (the Betti number).
Starting with $H_n(Y,Z)$, what we get is $$b_n(Y,Z) - b_n(X,Z) + b_n(X,Y) - b_{n-1}(Y,Z) + b_{n-1}(X,Z) - \cdots + (-1)^{n+1} b_0(X,Z).$$
The point is that if you collect all the terms corresponding to a fixed pair, you find that this gives $$(-1)^{n+1}\left(\chi(Y,Z) - \chi(X,Z) + \chi(Y,Z)\right) = 0.$$ Cancelling out the sign out front and moving sides around, you find the desired equality $\chi(X,Z) = \chi(X,Y) + \chi(Y,Z).$