Suppose we have a manifold W and on it a vector field. Is the manifold invariant under the vector field exactly if the vector field lies in the tangent space of W at w?
The tangent vectors are the vectors which are invariant?
Disclaimer: I know this question is probably much too vague.
The answer to the first question would be yes if you are thinking of $W$ as an embedded submanifold. Let's make this more precise. For simplicity, I will restrict to smooth manifolds, vector fields and so on.
In our setting, let $M$ be the ambient smooth manifold of dimension $m$ and $W$ the embedded submanifold $W$ of dimension $d$. Now consider a vector field $X$ on $M$ such that for each $w\in W$, $$X_w\in T_wW$$ which is the assumption that each vector of your vector field is tangent to the submanifold.
We have to show that for the local flow $(t,p)\mapsto \Phi(t,p)=:\Phi^t(p)$ of this vector field $X$ the following is satisfied:
Recall that the local flow is just a map such that 1) $\Phi(0,p)=p$ and 2) $\frac{d}{dt}\big|_{t=s}\Phi^t(p)=X_{\Phi^s(p)}$ for all $(s,p)$ in the domain $\mathcal{D}$, which is supposed to be an open neighbourhood of $\{0\}\times M$ in $\mathbb{R}\times M$ satisfying usually other sets of technical assumptions which are not important for us.
By the constant rank theorem, around every $w_0\in W$, there is a local chart of coordinates $(U_{w_0},(x^i))$ in $M$ in which $w_0=(0,\ldots,0)$ and in which $$W\cap U_{w_0}=\{x\mid x^{d+1}=x^{d+2}=\cdots=x^m=0\}.$$ In this local chart, for $w\in W\cap U_{w_0}$, the vector field $X$ will have the form $$X_w=\sum_{i=1}^da^i(x)\frac{\partial}{\partial x^i}.$$ But then for all $w\in W\cap U_{w_0}$ and $j>d$, the map $$t\mapsto x^j(\Phi^t(w))$$ is constant equal to zero as, by the chain rule, $$x^j(\Phi^t(w))'=\left(\sum_{i=1}^da^i(\Phi^t(w))\frac{\partial}{\partial x^i}\right)x^j=0$$ since $\frac{\partial}{\partial x^i}x^j=0$ for $j\neq i$. Hence the local flow satisfy the desired property as for all $w\in W\cap U_{w_0}$ and $t$, $\Phi^t(w)\in W$ since for all $j>i$, $$x^j(\Phi^t(w))=0.$$
Note 1: The above can be shown using other methods different from the constant rank theorem.
Note 2: A related theorem to look is Fröbenius theorem.
The answer to the second question would be yes in many formulation, as we can see the vector field $X$ and its flow as defined on $W$. Although, for a more concrete answer, you should clarify how you view the vector field acting $X$ acting on vectors.