First problem - (my original question before the editing)
Prove or disprove the following:
Let $A$, $B$ be differentiable manifolds such that $A \subseteq B$, and $s: A \to B$ a smooth map. Then $s \sim i$ where $i: A \to B$ is the inclusion map. The symbol "$\sim$" denotes "smoothly homotopic".
As Ben A. and jflipp pointed out, the claim is false.
Second problem
What if we add the hypothesis that $A$ and $B$ share the same homotopy type?
On
Your new claim is still false. If you take $A=B$, you're asking whether or not every map from $A$ to itself is homotopic to the identity; in particular this would imply the identity map is homotopic to a constant, and so $A$ is contractible. This is a very strong requirement to put on a manifold - most are far from it! It implies that all homology and homotopy groups are trivial, for one thing - and that $A$ was connected. The two-point space, the spheres, the tori, real projective spaces... all are not contractible. That this very special case fails to hold should show that your general hope is far from true.
Well, it can't work, because the claim simply is false, as pointed out by flip in the comment. If you want to know where the mistake is: the definition of the homotopy doesn't make sense in the second case ($0\leq t\leq 1$), because we neither know how to add elements in $B$, nor do we know what scalar multiplication is.