In Weintraub's Differential Forms, a complement to vector calculus, the following (Proposition 3.16) is given: Let $k(t) = f(t), g(t), h(t))$ be a smooth curve. Then: $$ k_* (i) = (f\prime(t), g\prime(t), h\prime(t)) = k\prime(t)$$ Where $i$ is tangent to $\mathbb{R}$ at t.
Proof: We want, first, to find a curve in $\mathbb{R}$, parameterized by some variable $s$, say, which passes through the point $t$ of $\mathbb{R}$ when $s=0$ and which has $r(0) = t$ and $r\prime(0) = i$. Then $k_*(i) = (kr)\prime(0)$. But $kr(s) = k(r(s)) = k(s+t)$, So $$(kr)\prime (0) = {d \over ds} k(s+t) |_{s=0} = (f\prime(t), g\prime(t), h\prime(t))$$
As claimed.
This makes sense to me, assuming that $\prime$ indicates the derivative of the variables with respect to $s$, which is being used to parameterize a curve on the manifold. In other words, if the subscript $_\gamma$ in some $f_\gamma$ denotes taking the derivative of $f$ with respect to $_\gamma$, then our previous expression can be written out as:
$$(f\prime(t), g\prime(t), h\prime(t)) = (f_s(t), g_s(t), h_s(t)) $$
However, some of his next examples throw me off, particularly and importantly this one:
Let $k(u,v) = (f(u,v), g(u,v), h(u,v))$ be a smooth surface. Then $$k_*(i) = f_u (u,v), g_u (u,v), h_u(u,v)) = k_u(u,v)$$
Now, all he gives to explain this is the following statement:
Proof: To compute $k_*(i)$, we look at the curve through the point $(u, v)$ given by $r(s) = (u + s, v)$. The computation is then very similar to (Proposition 3.16).
However, If I interpreted (Proposition 3.16) correctly then I would expect the solution to instead be of the form: $$k_*(i) = f_s (u,v), g_s (u,v), h_s(u,v)) = k_s(u,v)$$ As an aside, he also stated that: $$ k_*(j) = f_v (u,v), g_v (u,v), h_v(u,v)) = k_v(u,v)$$ However, I simplified this example, as I expect there to be no loss of generality. Thank you in advance for your help.