Proof that $P\subseteq NP$ without nondeterministic TM
The first definition you are referring to is probably the one that defines $\mathsf{NP}$ as the set of problems $\Pi$ for which there exist a non-deterministic poly-time Turing machine that decides $\Pi$.
Since any problem $\Pi \in \mathsf{P}$ can be decided by a deterministic poly-time Turing machine $T$, and $T$ itself is also a non-deterministic Turing machine, it follows that $\Pi \in \mathsf{NP}$, i.e., $\mathsf{P} \subseteq \mathsf{NP}$.
However $\mathsf{NP}$ can be equivalently defined as the set of decision problems $\Pi$ for which there exists a poly-time verifier (a Turing machine) $V$ and a polynomial $p$ such that if $x \in \{0,1\}^*$ is an instance of $\Pi$, then there exists a certificate $c \in \{0,1\}^*$ of length $|c| \le p(|x|)$ such that $V(x,c)$ accepts if and only if $x$ is a yes-instance for $\Pi$.
Using this definition every problem $\Pi \in \mathsf{P}$ is also in $\mathsf{NP}$, since it suffices to choose $p(|x|)=0$, $c(x)=\varepsilon$ (where $\varepsilon$ denotes the empty word), and $V(x, c)$ as the Turing machine that simulates any poly-time decider $T$ for $\Pi$ and accepts iff $T(x)$ accepts.