Q:

Why does curve25519 use a cofactor of 8? 

This cofactor (as I understand it) effectively discards valid points that satisfy the curve equation over the finite field. 
Why would one wish to reduce the number of possible private keys, it seems to reduce security. 
What is the reasoning behind using a cofactor of 8? 

A:

Curve25519 was chosen to have the Montgomery shape $y^2 = x^3 + A x^2 + x$ to support the fast single-coordinate Montgomery ladder for Diffie–Hellman: given $x(P)$ and $a$, it is cheap to compute $x([a]P)$, so there is no need to pass the $y$ coordinate and implementors are not tempted to use secret-dependent conditionals to compute scalar multiplication with the naive double-and-add algorithm—that is, the shape of Curve25519 was chosen to encourage implementors of the X25519 DH function to avoid timing side channels without sacrificing performance. 
Every Montgomery curve has a point of order 2 (if the equation holds for $P = (x, y)$ then it holds for $(x, -y) = -P$ so $P + (-P) = \mathcal O$), so the cofactor is always greater than 1—and when the coordinate field is $\mathbb Z/p\mathbb Z$ for $p \equiv 1 \pmod 4$ as is the case for $2^{255} - 19$, the orders of the curve and its twist are 4 and 8 or vice versa. 
Curve25519 was chosen to minimize the cofactors, and when taking all the other criteria into account, the best choice had cofactor 8 for the curve and cofactor 4 for the twist.   RFC 7748 gives a deterministic algorithm to search according to these criteria in Appendix A.  For comparison, the Curve448 coordinate field is modulo $p = 2^{448} - 2^{224} - 1$ with $p \equiv 3 \pmod 4$, so both the curve and its twist have cofactor 4.