The definition of a knot The definition of a knot is an injective piecewise linear map from $S^{1} $ to $\mathbb{R}^{3} $. Isn't that equivalent to a subset of $\mathbb{R}^{3} $ homeomorphic to $S^{1} $ that is piecewise linear?
 A: Let us rephrase your question in the following way:
"The definition of a knot is an embedding of $S^1$ in $\mathbb{R}^3$. Isn't that equivalent to a submanifold of $\mathbb{R}^3$ isomorphic to $S^1$?"
The answer is no. A submanifold can be embedded in different ways: Compose a given embedding with any isomorphism of $S^1$ that isn't the identity and you get another embedding.
Note however, that two knots $a, b: S^1 \to \mathbb{R}^3$ are being considered equivalent (with some mathematicians even equal) if there is an isotopy $H: \mathbb{R}^3 \times [0, 1] \to \mathbb{R}^3$, such that $a = H_1 \circ b$.
We could try to recreate this notion of equivalence for your second definition:
Two "knots" $a, b \subset S^1$ are equivalent if there is an isotopy $H: \mathbb{R}^3 \times [0, 1] \to \mathbb{R}^3$, such that $H_1(a) = b$.
So if two knots are equivalent in the first definition, they are equivalent in the second definition. I'm not sure if the converse is true, i.e. whether there is an ambient isotopy for every automorphism of the embedded manifold, but I would believe so. So if you consider equivalence classes of knots, it shouldn't matter which definition you use, although I recommend the first one.
Comments:


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*I rephrased your question such that it makes sense in any category of manifolds, e.g. piecewise linear, smooth, analytical, topological or whatever you want. If your question was about technical aspects of piecewise linear categories, please rephrase it accordingly.

*Often, knots are considered as embeddings of $S^1$ into $S^3$, which is the one-point-compactification of  $\mathbb{R}^3$. My answer applies to that definition equally.

