Let $M$ be a $n$-dimensional manifold and $S \in M$ a $d$-dimensional submanifold. We will assume the manifolds are real manifolds.
Let $p \in S$, $m=n-d$ and $F=(f_1,..,f_m): M \longrightarrow \mathbb{R}^m$ be defined and submersive at $p$ and such that $S$ is defined in a neighbourhood of $p$ by the system $$ f_i=0 \qquad i=1,..,m$$
This implies that $T_p S$= Ker $T_p f$. The proof of this fact in my course notes just states it is an immediate consequence of the implicit function theorem and the definition of submersive maps. However, I don't see if I can see that clearly.
For me, this seems to follow from the fact that dim(Im $T_p f$)=$m$ as $f$ is submersive, implying the dimension of Ker $T_p f$ is $d$ which is the same as that of $T_p S$ and therefore these two vector spaces are equal up to isomorphism.
Am I reasoning correctly?
No. The statement is that the spaces $T_pS$ and $\text{Ker} \,T_pf$ are equal, not that they're isomorphic. There's a big difference. (For example the $xy$-plane and the $yz$-plane in $\mathbb R^3$ are isomorphic because they have the same dimension. But if you thought they were equal, you wouldn't know whether you're standing up or lying down!)
You need to observe first that the definition implies that $T_pS \subseteq \text{Ker}\, T_pf$. Then you can argue that they have the same dimension, and so they must be equal.