Let $Y$ be a smooth projective variety, $\omega_Y$ its canonical bundle and $k(y)$ the skyscraper sheaf. In a proof in Huybrechts' book, he uses the "restriction" map $r_{y_1,y_2} \colon \omega_Y^k \to \omega_Y^k(y_1) \oplus \omega_Y^k(y_2) \simeq k(y_1) \oplus k(y_2)$, where $y_1$ and $y_2$ are distinct closed points.
I don't understand how the map is defined and why the isomorphism $\omega_Y^k(y_1) \oplus \omega_Y^k(y_2) \simeq k(y_1) \oplus k(y_2)$ holds.
For any affine scheme $X:=Spec(A)$ and any $A$-module $E$ there is a canonical map
$$ \rho_E(x): E \rightarrow E\otimes_A \kappa(x)$$
defined by $\rho_E(x)(e):= e\otimes 1$, where $1\in \kappa(x)$ is the multiplicative identity. Here $\kappa(x)$ is the residue field of the point $x$: If you let $\mathfrak{p}\subseteq A$ be the prime ideal corresponding to $x$ you define $\kappa(x):=A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}}$. It follows $\rho_E(x)$ is a map of $A$-modules, hence you get an induced map of $\mathcal{O}_X$-modules
$$\rho_{\mathcal{E},x}: \mathcal{E} \rightarrow \mathcal{E}(x).$$
If $E:=A$ you get a map
$$\rho_{\mathcal{O}_X,x}: \mathcal{O}_X \rightarrow \underline{\kappa(x)}.$$
The sheaf $\underline{\kappa(x)}$ is the sky-scraper sheaf of the point $x$ (this is defined in Hartshornes book).
If $x,y \in X$ are two points there is a canonical map
$$\rho_{\mathcal{O}_X,(x,y)}: \mathcal{O}_X \rightarrow \underline{\kappa(x)\oplus \kappa(y)}$$
defined using the canonical map
$$ \rho_A(x,y): A \rightarrow \kappa(x)\oplus \kappa(y),$$
where $\rho_A(x,y)(a):=(\rho(x)(a), \rho(y)(a))\in \kappa(x)\oplus \kappa(y).$
Example. If $Y$ is a smooth complex affine algebraic variety of dimension $n$ and $x\in Y$ a (closed) point, it follows the map
$$\rho_{\mathcal{O}_Y,x}: \mathcal{O}_Y \rightarrow \underline{\kappa(x)}$$
evlauates a global section at the point $x$. If $Y:=Spec(A)$ where $A$ is a finitely generated algebra over the complex numbers, and $f\in A$ is a global section of the structure sheaf, it follows the canonical map
$$\rho_{A,x}: A \rightarrow \kappa(x)$$
is the evaluation: If $x\in Y \subseteq \mathbb{C}^n$ where $x:=(x_1,..,x_n)$ it follows $\rho_{A,x}(f):=f(x_1,..,x_n)$ is the value of the function $f$ at $x$. In algebraic geometry you sometimes meet non-closed points, hence the point $x$ may correspond to a non-maximal prime ideal $\mathfrak{p}$ in $A$, and then the "value" of $f$ at $x$ becomes more un-intuitive. It is an element in the residue field $\kappa(x)$ of $x$, and this field is a transcendental extension of the base field.
When the base field $k$ is non-algebraically closed and $Y$ is a scheme of finite type over $k$ it follows from the HNSS the residue field $\kappa(x)$ at a closed point is a finite extension of the base field $k$. The value $s(x)$ of a global section lives in the residue field $\kappa(x)$ and this field changes with the point. If $k$ is the ring of integers and $x,y$ are two closed points in $Y$ it follows the residue fields $\kappa(x)$ and $\kappa(y)$ may have different characteristic.
If $Y$ is smooth and we define the canonical bundle as follows: $\omega_Y:=\wedge^n \Omega^1_Y$ (see Hartshorne, page 180), and since the rank of the cotangent bundle is $n$, it follows $\omega_Y$ is an invertible sheaf on $Y$. Hence it is locally trivial of rank one, hence you get something similar: There is a map
$$I0,\text{ } \rho_{\omega, x}: \omega_Y \rightarrow \underline{\kappa(x)}$$
Question: "I don't understand how the map is defined and why the isomorphism $ω_Y(y_1)\oplus ω_Y(y_2)\cong k(y_1)\oplus k(y_2)$ holds".
Answer: Since the rank of $\omega_Y$ is one, it follows the fiber $\omega_Y(x)$ is isomorphic to $\kappa(x)$ - the fiber is one-dimensional, hence there is in the case when $Y$ is smooth an isomorphism
$$ \omega_Y(x) \oplus \omega_Y(y) \cong \kappa(x) \oplus \kappa(y).$$
If $\omega^k_Y:=\omega^{\otimes k}_Y$ is the $k$'th tensor product of $\omega_Y$ with itself it follows $\omega^k_Y$ is still an invertible sheaf, and you get a similarly defined isomorphism
$$I1. \text{ } \omega^k_Y(x) \oplus \omega^k_Y(y) \cong \kappa(x)\oplus \kappa(y).$$
Note: The isomorphisms in $I0$ and $I1$ depend on the choice of a local trivialization of $\omega^k_Y$ around the points $x,y$.
Example. If $k:=\mathbb{R}$ is the real numbers and $A:=\mathbb{k}[x]$ is the ring of polynomials with real coefficients you may classify all maximal ideals. A maximal ideal $\mathfrak{m}$ in $A$ on the following form: If the residue field is the field of real numbers $k$ it follows $\mathfrak{m}_1=(x-r)$ with $r\in k$ a real number. If the residue field is the field $K:=\mathbb{C}$ of complex numbers, it follows $\mathfrak{m}_2$ is on the form $\mathfrak{m}_2=(p_z(x))$ where $z:=a+ib \in K-k$ is a non real complex number and
$$p_z(x):=(x-z)(x-\overline{z})=x^2-2ax+a^2+b^2.$$
In the case $1$ you get an evaluation map
$$\rho_1: A \rightarrow \kappa(\mathfrak{m_1})$$
defined by $\rho_1(f(x)):=f(r)$. In the case $2$ you get an evaluation map
$$\rho_2: A \rightarrow \kappa(\mathfrak{m}_2)$$
and if you want an explicit complex number you must choose an isomorphism $i:\kappa(\mathfrak{m}_2)\cong \mathbb{C}$. Here you have two choices: $i(x):=z$ or $i(x):=\overline{z}$. Once you have made a choice you get $\rho_2(f(x)):=f(i(x))\in \mathbb{C}$. Hence for a point that is not rational over $k$ your "value" and "evaluation map" depends on the choice of an isomorphism $i$. I believe this is written out in detail in Hartshornes book "Algebraic geometry".
When doing algebraic geometry you are seldom interested in the "value" of a function $f(x)$ or a section $s(x)$ in the fiber. What you are studying is where the function (or section) is zero, and this does not depend on the choice of an isomorphism $i$. Hence the fact that you have to choose an isomorphism $i$ above does not create problems.
Some people argue that to introduce non-maximal prime ideals is "less intuitive", but as you can observe above: If you work over a non-algebraically closed field, the elements in the structure sheaf do not give rise to functions in the classical sense.
