Let $R$ be a local ring with maximal ideal $\mathfrak{m}$ and let $(X, \mathcal{O}_X)$ be a scheme.
I want to show that there is a bijection:
$$ \begin{align} & \quad \{ (f, f^\#) \in \mathrm{Hom}(\operatorname{Spec}(R), X) \mid f(\mathfrak{m}) = x \} \\ \longleftrightarrow & \quad \mathrm{Hom}_{\text{local ring}}(\mathcal{O}_{X, x}, R)\\ \end{align} $$
So far, I have found maps in both directions (detailed below), but am stuck on showing that these are injective (either directly, or by showing that they are mutual inverses).
Specifically, given $(f, f^\#)$, we have an induced local ring homomorphism, $f^\#_\mathfrak{m}: \mathcal{O}_{X, x} \rightarrow \mathcal{O}_{\operatorname{Spec}(R), \mathfrak{m}} \cong R_{\mathfrak{m}} \cong R$.
In the other direction (for the case that $X = \operatorname{Spec}(S)$ is an affine scheme), suppose we have $\varphi: \mathcal{O}_{X, x} \cong S_x \rightarrow R$. Since this is a local homomorphism, $\varphi^{-1}(\mathfrak{m}) = x \cdot S_x$, the maximal ideal in $S_x$. This induces a homomorphism
$$ S \xrightarrow{\ell_x = \text{localization}} S_x \overset{\varphi}{\longrightarrow} R $$
and hence, $(\varphi \circ \ell_x)^{-1}$ is a map from $\operatorname{Spec}(R)$ to $\operatorname{Spec}(S) = X$ with $(\varphi \circ \ell_x)^{-1}(\mathfrak{m}) = x$, as required .
[And in the case that $X$ is not affine, we just do the same thing with an affine neighbourhood of $x$. Reducing from the general case to the affine case seems to be straightforward when showing injectivity as well...]
What I am struggling with is showing that the above maps are indeed injective (or, even better, mutually inverse) even in just the affine case. I have just started learning about schemes, so my knowledge is a bit limited, and whilst more abstract explanations may be useful, I might find more 'concrete' explanations a bit easier to understand.
From the map $g :=(\varphi \circ l_x)^{-1}: \mbox{Spec}(R) \to \mbox{Spec} (S)$ we know that $g^{\#}(\mbox{Spec} (S))= \varphi \circ l_x$, so we have the induce map $g^{\#}_x : S_x \to R_{\mathfrak{m}} = R$ which is just $ (\varphi \circ l_x)_{x} = \varphi$.
Let $(f,f^{\#}) : \mbox{Spec}(R) \to \mbox{Spec} (S)$ and from your construction we have the local morphism $f^{\#}_{\mathfrak{m}} : S_x \to R $. Now we can construct $g = (f^{\#}_{\mathfrak{m}} \circ l_x)^{-1} : \mbox{Spec}(R) \to \mbox{Spec} (S)$. We know that $f= f^{\#}(\mbox{Spec} (S))^{-1}$, so we have for every $\mathfrak{p} \in \mbox{Spec} (R)$, $(f^{\#}_{\mathfrak{m}})^{-1}(\mathfrak{p}) = f(\mathfrak{p}) S_x$. From this $g = f$.
So the maps that you construct are mutually inverse.