Let $X$ be a regular projective surface over a field, and consider the inclusions: $x\in Y\subset X$ where $x$ is a closed point and $Y$ is an integral curve on $X$. Let $k(x)=\frac{\mathcal O_{X,x}}{\mathfrak m_x}$ be the residue field at $x$. Moreover let $\mathfrak m _{x,y}$ be the maximal ideal of $\mathcal O_{Y,x}$ and consider $k'(x):=\frac{\mathcal O_{Y,x}}{\mathfrak m_{x,y}}$. I other words $k'(x)$ is the residue field at $x$ when considered as a point on the curve $y$.
Is it true that $k(x)=k'(x)$?
Not only is what you say true, but it is true under the most general conceivable situation: no base field, no regularity assumption, no projective assumption, no curve assumption ... More precisely:
Claim:
Let $X$ be a scheme, $Y\subset X$ a closed subscheme and $x\in Y$ a closed point.
Then the canonical morphism $k_X(x)\stackrel {\cong}{ \to} k_Y(x)$ is an isomorphism of fields.
Proof:
By taking an open affine neighbourhood of $x$ in $X$ we may reduce to the case where $X=\operatorname {Spec} A$, $x=\mathfrak m\subset A $ a maximal ideal and $Y=V(I)$ the closed subscheme corresponding to an ideal $I\subset \mathfrak m$. The canonical map of residue fields is then the isomorphism $$k_X(x)=\frac {A}{\mathfrak m}\stackrel {\cong}{ \to} k_Y(x)=\frac {A/I}{{\frak m}/I}$$