When one wants to know that whether a partial function $f \colon \Sigma^{*} \supsetneq \mathrm{dom}(f) \rightarrow \Sigma^{*}$ is Turing-computable, there are two methods that I think they are both useful.
We can define a total function $\bar{f}$ such that $$ \bar{f}(x) = \left\{ \begin{aligned} &f(x),&x \in \mathrm{dom}(f) \\ &\bot,&x \not\in \mathrm{dom}(f) \\ \end{aligned} \right.$$ Thus, $f$ is Turing-computable if $\bar{f}$ is Turing-computable by a TM using $\bar{\Sigma} = \Sigma \cup \{ \bot \}$. (Some answers use this definition. see [1] and [2])
Let $M$ be a TM, and We say $f$ is Turing-computable by $M$ if $$M(x) = \left\{ \begin{aligned} &f(x),&& x \in \mathrm{dom}(f) \\ &\bot,&&x \not\in \mathrm{dom}(f) \\ \end{aligned} \right.$$ where $M(x) = \bot$ means that $M$ never halts on $x$. (This definition is introduced in my books.)
These two definitions are not equivalent, since $M$ can computes some functions whose domain is Turing-recognizable instead of Turing-decidable in sense of definition 2.
I want know that is the definition 2 more powerful? In another word, if $f$ is Turing-computable in sense of definition 1, can we prove that it is also Turing-computable in sense of definition 2?