Here is a quote from the book Algebraic Topology by Allen Hatcher:
Here is an interesting application of degree:
Theorem 2.28 $S^n$ has a continuous field of nonzero tangent vectors iff $n$ is odd
Proof: Suppose $x \mapsto v(x)$ is a tangent vector field on $S^n$, assigning to a vector $x \in S^n$ the vector $v(x)$ tangent to $S^n$ at $x$. Regarding $v(x)$ as a vector at the origin instead of at $x$, tangency just means that $x$ and $v(x)$ are orthogonal in $\mathbb{R}^{n+1}$. If $v(x) \neq 0$ for all $x$, we may normalize so that $|v(x)| = 1$ for all $x$ by replacing $v(x)$ by $v(x) / |v(x)|$. Assuming this has been done, the vectors $(\cos t)x + (\sin t)v(x)$ lie in the unit circle in the plane spanned by $x$ and $v(x)$. Letting $t$ go from $0$ to $\pi$, we obtain a homotopy$f_t(x) = (\cos t)x + (\sin t)v(x)$ from the identity map of $S^n$ to the antipodal map $- \mathbb{1}$. This implies that $\text{deg}(-\mathbb{1}) = \text{deg}(\mathbb{1})$, hence $(-1)^{n+1} = 1$ and $n$ must be odd.
Conversely, if $n$ is odd, say $n = 2k-1$, we can define $v(x_1, x_2, \cdots, x_{2k-1}, x_{2k}) = (-x_2, x_1, \cdots, -x_{2k}, x_{2k-1})$. Then $v(x)$ is orthogonal to $x$, so $v$ is a tangent vector field on $S^n$ and $|v(x)| = 1$ for all $x \in S^n$.
In the proof of Theorem 2.28, I am not getting the need of $x$ and $v(x)$ being orthogonal unit vectors. What if they are not orthogonal to each other?
The fact that $v(x)$ is a tangent vector to $S^n$ at $x$ implies that $v(x)$ is orthogonal to $x$. The fact that $v(x) \neq 0$ for all $x$ means that we can assume without loss of generality that the $v(x)$'s are unit vectors for all $x$, for if they are not to begin with, then you can rescale, as described in the book.
For the remainder of the proof, what's really important is that:
The proof of this requires $v(x)$ to be orthogonal to $x$ (i.e. $v(x).x = 0$), and also requires $v(x)$ to be a unit vector (i.e. $|v(x)|^2 = 1$), for all $x \in S^n$.
Essentially, to prove that the vector $(\cos t) x + (\sin t) v(x)$ lies on $S^n$, you have to show that it has unit norm. The following calculation does precisely this. \begin{align*} \left| (\cos t) x + (\sin t) v(x) \right|^2 & = (\cos^2 t) |x|^2 + (2\cos t \sin t) (x.v(x)) + (\sin^2t) |v(x)|^2 \\ & = \cos^2 t+\sin^2 t \\ & = 1,\end{align*}
Having established that $(\cos t) x + (\sin t) v(x)$ lies on $S^n$, it makes sense to define a function $$ F: [0, \pi ] \times S^n \to S^n$$ using the formula $$ F(t, x) = (\cos t) x + (\sin t) v(x).$$
This $F$ is a homotopy from the identity map $f(x) = x$ to the antipodal map $g(x) = x$. When $n$ is even, no such homotopy can exist since $\text{deg}(f) \neq \text{deg}(g)$. This contradiction proves that the non-vanishing vector field $v$ cannot exist.