Let $M$ be a $7$-dimentional manifold with nonempty boundary $\partial M$ and let $f:M\rightarrow N$ be a smooth mapping where $N$ is $5$-dimentional. Assume tham for some $p\in\partial M$ we have $rank(f|_{\partial M},p)=3$. I need to prove that $$3\leq rank(f,p)\leq4$$ I think I have the left side figured out. We know that $\partial M$ is $6$-dimentional manifold. The matrix of derivatives has $6$ columns and $5$ rows. We know that rank of that matrix is $3$. On the other hand when we consider a whole manifold $M$, then we have a matrix with $7$ columns and $5$ rows. Rank of such matrix cannot be less than $3$ because it has the matrix for $\partial M$ in it.
Since $N$ is $5$-dimetional we have $rank(f,p)\leq 5$ but I don't know how to prove the right inequality.
Suppose you have $\phi: U \to V \subset \mathbb{H}^7$ is a chart around $p \in U$ that sends $p$ to $0$ and $\partial M$ to the $x_7 = 0$ hyperplane i.e. you have a boundary chart around $p$. Then $$3 = \text{rank} (f|_{\partial M},p) = \text{rank} D(\hat{f} \circ i)_p$$
(where $i$ is the inclusion of the hyperplane into $\mathbb{H}^7$, and $\hat{f} = f \circ \phi^{-1}$ is the coordinate representation of $f$) which as you point out is just looking at $D\hat{f}_p$ in coordinates and omitting the bottom row. Since the rank of this $6 \times 5$ matrix is $3$, that means $3$ of the rows are linearly independent. Now looking at $D\hat{f}_p$ as a whole, this just adds another row vector underneath, so you have two cases. If there are still only $3$ linearly independent rows, then $\text{rank} D\hat{f}_p = 3$, else if the new row is linearly independent of the other vectors, then $\text{rank} D\hat{f}_p = 4$. So $3 \leq \text{rank} D\hat{f}_p \leq 4$ and so $3 \leq \text{rank} (f,p) \leq 4$.