Definiton A $p$-adic integer is a (formal) series $$\alpha=a_0+a_1p+a_2p^2+\ldots$$ with $0\leq a_i<p$.
The set of $p$-adic integers is denoted by $\mathbb{Z}_p$. If we cut an element $\alpha\in\mathbb{Z}_p$ at its $k$-th term $$\alpha_k=a_0+a_1p+\ldots+\alpha_{k-1}p^{k-1}$$ $\textbf{we get a well defined element of }$ $\mathbf{\mathbb{Z}/p^k\mathbb{Z}}$.
Could someone explain me the bold part?
On
From your comment to Hurkyl I reckon that the following answers your intended question:
In our decimal number system we can write numbers $0\leq x < 10^k$ in the form $$ 10^{k-1} a_{k-1}+10^{k-2}a_{k-2}+...+10 a_1+a_0 $$ with $0\leq a_i<9$ being the digits of that number. Pick a specific $k$ and think this through and it should be clear. The same system works in base $p$ so that $$ p^{k-1} a_{k-1}+p^{k-2}a_{k-2}+...+p a_1+a_0 $$ defines a unique number $0\leq x<p^k$. In fact you can just check that the largest number that can be written in this form is if all $a_i=p-1$ which then yields $$ \begin{align} &p^{k-1}(p-1)+p^{k-2}(p-1)...+p(p-1)+(p-1)\\ =&p^k-p^{k-1}+p^{k-1}-p^{k-2}+...+p^2-p+p-1\\ =&p^k-1 \end{align} $$ which is in fact the largest integer satisfying $0\leq x<p^k$. The above sum is known as a telescoping sum since all but the first and last term cancels out.
In $\Bbb Z[[X]]$ say we have two formal power series $P(X)$ and $Q(X)$. Suppose their "truncations" up to the point $X^n$ are $p(X)$ and $q(X)$. That is, $P(X)=p(X)+X^nA(X)$ and $Q(X)=q(X)+X^nB(X)$ for some formal power series $A(X),B(X)\in\Bbb Z[[X]]$ and polynomials $p(X),q(X)$ of degrees $<n$; it is best to think of $p$ and $q$ as the remainders of $P$ and $Q$ mod $X^n$. Then
$$P(X)Q(X)=(p+X^nA)(q+X^nB)=p(X)q(X)+X^n[A(X)+B(X)+X^nA(X)B(X)].$$
Thus, truncating $P(X)$ and $Q(X)$ up to $X^n$ and multiplying them together can be done in any order. Mentally, this tells us that the truncations act as elements of $\Bbb Z[[X]]/(X^n)$, and this is in fact true: the taking a series $P(X)$ to the coset $q(X)+(X^n)$ is a quotient map $\Bbb Z[[X]]\to\Bbb Z[[X]]/(X^n)$.
The same exact idea is in play with $\Bbb Z_p$. We have a projection $\Bbb Z_p\to\Bbb Z/p^n\Bbb Z$ that takes a $p$-adic expansion to the coset with representative given by the truncation of that series. This map is a well-defined ring homomorphism with kernel $p^n\Bbb Z_p$ so $\Bbb Z_p/p^n\Bbb Z_p\cong\Bbb Z/p^n\Bbb Z$.
It would make more sense to use the phrase "well-defined" in the context of more general $p$-adic expansions; in general the coefficients need not be integers $0\le a_i\le p-1$, they could be any integers (even other $p$-adic integers; any system of coset representatives for $p\Bbb Z_p$ would work as a set of "digits" $-$ this is usually introduced as an exercise to those first learning about the $p$-adics).
The fact that the "truncate the power series" process gives a unique image of a $p$-adic integer under the projection map regardless of how the power series is defined is what makes the truncation expression as an element of $\Bbb Z/p^n\Bbb Z$ well-defined (independent of how the $x\in\Bbb Z_p$ is represented).