Why are the total functions not enumerable?
Because of diagonalization. If $(f_e: e \in \mathbb{N})$ was a computable enumeration of all total computable functions from $\mathbb{N}$ to $\mathbb{N}$, such that every $f_e$ was total, then $g(i) = f_i(i)+ 1$ would also be a total computable function, but it would not be in the enumeration. That would contradict the assumptions about the sequence. Thus no computable enumeration of functions can consist of exactly the total computable functions. 

Suppose we think of a universal computable function $h(e,i)$, where "universal" means $h$ is a computable binary function and that for every total computable unary function $f(n)$ there is some $e$ such that $f(i) = h(e,i)$ for all $i$. Then there must also be some $e$ such that $g(n) = h(e,n)$ is not a total function, because of the previous paragraph. Otherwise $h$ would give a computable enumeration of total computable unary functions that includes all the total computable unary functions. 

Thus the requirement that every function is a system of functions is total is incompatible with the existence of a universal function in that system. For some weak systems, such as the primitive recursive functions, every function is total but there are not universal functions. Stronger systems that have universal functions, such as Turing computability, simply must have partial functions in order to allow the universal function to exist.