Clarifying the definition of reduction with regards to NP-complete problems
A problem $L$ is $\text{NP}$-complete if $L$ is in $\text{NP}$, and $L$ is $\text{NP}$-hard (that is, $A\leq_p L$ for all $A\in \text{NP}$ ). Consider the following claims.
Claim 1: if $L$ is $\text{NP}$-complete and $L\in \text{P}$, then $\text{NP} \subseteq \text{P}$ (that is, all problems in $\text{NP}$ can be solved in deterministic polynomial time).
Claim 2: If $ A \leq_p B$, then $B\in \text{P} \to A \in \text{P}$.
What you asked about is claim 1. To begin with, its correctness follows from claim 2. Indeed, if $L$ is $\text{NP}$-complete, then for every $A\in \text{NP}$, it holds that $A\leq_p L$. Thus, if we assume that $L\in \text{P}$, then we get by claim 2 that $A\in \text{P}$, and therefore $\text{NP} \subseteq \text{P}$. So what is missing for you is the proof of claim 2, and a sketch of its proof goes as follows. If $M_f$ is a TM that computes a polynomial time reduction $f$ from $A$ to $B$, and $M_B$ is a TM that decides $B$ in polynomial time, then a TM $M_A$ that decides $A$ in polynomial time operates as follows. On input $x$, $M_A$ computes $y = M_f(x)$, then $M_A$ runs $M_B$ on $y$, and answers the same. What is left to show is that $M_A$ decides $A$ in polynomial time, and I leave that to you (use the fact that |y| is polynomial, and the fact that a composition of polynomials is a polynomial).