To answer your first question, i.e., what is type on a (polarized) abelian variety $A$? It is already answered here. Roughly speaking, for a polarized abelian variety $(A,L)$, there is an integral basis $\{dx_i,dy_i\}_{i=1}^{g}$ of $H^1(A,\mathbb Z)$, and positive integers $d_1,\ldots, d_g$ with $d_i|d_{i+1}$ such that
$$c_1(L)=\sum_i d_idx_i\wedge dy_i.\tag{1}\label{1}$$
The $g$-tuple $(d_1,\ldots,d_g)$ is called the type of $(A,L)$.
For example, when $A=J(C)$ is the Jacobian variety of a smooth projective curve $C$, one can choose $L$ to be principal polarization ($d_1=\cdots=d_g=1$). This follows from the fact that there is a symplectic basis $\{\gamma_i,\delta_i\}_{i=1}^g$ on $H_1(J(C),\mathbb Z)=H_1(C,\mathbb Z)$ such that $\gamma_i\cdot\delta_j=\delta_{ij}$ and $\gamma_i\cdot\gamma_j=\delta_i\cdot\delta_j=0$.
For your second question, in fact, the book (Birkenhake-Lange, p.119) defines the degree of a polarization of $L$ to be the product $\prod_id_i$. So the relation you're looking for is probably
$$\chi(L)=\prod_id_i.\tag{2}\label{2}$$
Proof of \eqref{2}. First, by Hirzeburch-Riemann-Roch theorem, one obtains
$$\chi(L)=\int_Ach(L)td(A)=\int_Ach(L)=\int_A\frac{c_1(L)^g}{g!},\tag{3}\label{3}$$
where the second and the third identities follow from the fact that the tangent bundle of $A$ is trivial and the definition of the Chern character on a line bundle.
Now, by \eqref{1}, one has
$$c_1(L)^g=(g!\prod_id_i)dx_1\wedge dy_1\wedge\cdots\wedge dx_g\wedge dy_g,$$ so $\int_Ac_1(L)^g=g!\prod_id_i$. Combine with identity \eqref{3}, the equality \eqref{2} follows. $\Box$
As a final remark, the proof of the identity \eqref{2} can be more concrete: By Kodaira vanishing theorem, all higher degree cohomology of $L$ vanish, so $\chi(L)=\dim H^0(A,L)$, one it reduces to show that there are exactly $\prod_id_i$ independent global sections on $L$. These sections correspond to theta functions on $A$. One can refer to Griffiths-Harris, p.317-320 for how these theta functions are found explicitly.
