Why are the random exponents so much bigger in the Socialist Millionaire protocol versus Diffie-Hellman key exchange?
For Diffie-Hellman, adequate security is achieved provided that:


we work modulo a prime $p$ big enough to resist discrete logarithm (1536 bits are sufficient);
the order of the subgroup generated by $g$ is a multiple of a big-enough prime integer $q$ ($q$ should have length $2n$ bits to achieve $2^n$ security);
the private exponents are randomly chosen in a big enough range ($2n$-bit exponents to reach $2^n$ security).


The first two conditions are necessary for the discrete logarithm to be hard. But having hard discrete logarithm is not necessarily sufficient for the security of the cryptographic algorithm which uses that group. Diffie-Hellman appears to be reasonably secure with the conditions above. However, other protocols require a bit more. A prime example is DSA. With DSA, you work in a subgroup of size $q$ (a prime integer) generated by a conventional value $g$ modulo a big prime $p$; whenever you sign, you select a random $k$ modulo $q$ (that's a "secret exponent"). $k$ must be chosen uniformly in the $[1...q-1)]$ range; simply choosing a "big enough" $k$ is not sufficient: any bias in the selection of $k$ can be exploited to rebuild the private key from the observation of several signatures (many signatures, if the bias is slight).

So your question is: is SMP more like Diffie-Hellman, i.e. tolerates that private exponents be chosen in a subrange, or more like DSA, for which uniform selection in the whole range is necessary ? Barring any explicit indication to the contrary, it is safer to assume that uniform selection in the whole range should be used. In particular, it seems to me that non-uniform selection may imply exploitable biases in the zero-knowledge proofs which are sent along the SMP values.