(Below, for simplicity all functions are unary. Also, "$\uparrow$" denotes "is undefined." Finally, for simplicity I'll think of languages as sets of natural numbers as opposed to sets of strings, although this is purely superficial.)
The key is the following "translation" from partial functions to sets:
Given a partial function $f$, let $$Graph_f=\{\langle a,b\rangle: f(a)=b\}$$ (where "$\langle\cdot,\cdot\rangle$" is your favorite pairing operation).
Note that $Graph_f$ is defined for all partial functions $f$, not just the recursive ones. The construction $f\mapsto Graph_f$ lets us translate properties of functions to corresponding properties of sets, and in general all computability-theoretic properties will "match up" appropriately. In particular, we have:
Suppose $f$ is a partial function. Then the following are equivalent: $(i)$ $f$ is a partial recursive function. $(ii)$ $Graph_f$ is recursively enumerable.
(Indeed, this is one way that partial recursive functions are sometimes defined in the first place. Moreover, note that in much of logic a function literally is its graph, so it's unsurprising to see this sort of correspondence.)
On a less-immediate note, observe that the further assumption of totality simplifies things substantially:
Suppose $f$ is a total function. Then the following are equivalent: $(i)$ $f$ is recursive. $(ii)$ $Graph_f$ is r.e. $(iii)$ $Graph_f$ is recursive.
Clearly $(iii)\rightarrow (ii)$, and $(ii)\rightarrow (i)$ by the previous observation. To see $(i)\rightarrow (iii)$, suppose $f$ is total recursive and we want to check whether $\langle a,b\rangle\in Graph_f$; we simply run (the computation of) $f$ on input $a$ until it halts and outputs some $n$, which it must since $f$ is total, and then we check whether $n=b$.
If we drop the totality assumption this breaks down: the function $$h_X(n)=\begin{cases}
n & \mbox{ if $n\in X$}\\
\uparrow & \mbox{ otherwise}\\
\end{cases}$$ is partial recursive whenever $X$ is r.e., but $Graph_{h_X}=\{\langle x,x\rangle: x\in X\}$ has the same Turing degree as $X$, so if $X$ is a non-recursive r.e. set then $h_X$ is a partial recursive function with r.e. but non-recursive graph.
It's also worth noting that there's also a translation "dual" to the $f\mapsto Graph_f$-construction, although it's a bit more finicky. Every r.e. set $X$ has a "one-at-a-time" enumerator - that is, a machine $M$ such that for each $s$ there is at most one $t_s$ such that $M$ accepts $t_s$ in exactly $s$ steps. The associated map $$g_M(s)=\begin{cases}
t_s & \mbox{ if $M$ accepts some string in exactly $s$ steps}\\
\uparrow & \mbox{ otherwise}
\end{cases}$$ is partial recursive and we have $Graph_{g_M}=X.$
The reason this is more finicky is that $M$ is not uniquely determined by $X$ - every r.e. $X$ will have many such $M$s.
