On heuristic introduction to tangent spaces, T.A.Springer on his linear algebraic group book (chapter 4,subsection 4.1.2) mentioned about partial derivations. I didn’t understand What did he mean by it and how he get an expression for $f_i(x+tv)$.
I attached the same as an image.
Kindly help me with this. 
thank you
Question: "On heuristic introduction to tangent spaces, T.A.Springer on his linear algebraic group book (chapter 4,subsection 4.1.2) mentioned about partial derivations. I didn’t understand What did he mean by it and how he get an expression for fi(x+tv)."
Example: If $f(T):=T^2$ you get
$$f(x+tv)=(x+tv)^2=x^2+2xtv+t^2v^2=x^2+ t(v(2x) +tv^2)=$$
$$f(x)+t(vD_x(f(x))+tv^2),$$
where
$$D_x:=\frac{\partial}{\partial_x}$$
is partial derivative wrto the $x$-variable. For any polynomial $f(x):=\sum_i a_i x^i$ you may define the "x-derivative"
$$D_x(f(x)):= \sum_i a_i ix^{i-1}$$
and prove that
$$L1.\text{ }D_x(f(x)g(x))=D_x(f(x))g(x)+f(x)D_x(g(x))$$
hence the operator $D_x:k[x] \rightarrow k[x]$
is a $k$-linear map verifying the rule $L1$. You may in general define for any $k$-algebra $A$, the $A$-module of derivations $Der_k(A)$. By definition, $Der_k(A)$ is the set of $k$-linear maps
$$\partial: A \rightarrow A$$
with $\partial(ab)=a\partial(b)+\partial(a)b$ for all $a,b \in A$. It follows $Der_k(A)$ is a left $A$-module and $k$-Lie algebra with Lie product $$[\partial, \partial']:=\partial \circ \partial' - \partial' \circ \partial$$
**Question: What is partial derivations?"
Answer: If $A:=k[x_1,..,x_n]$ is the polynomial ring in the variables $x_i$ it follows there is an operator $\partial_{x_i} \in Der_k(A)$ defined by
$$\partial_{x_i}(\sum_I a_Ix_1^{j_1}\cdots x_i^{j_i}\cdots x_n^{j_n}):=$$
$$\sum_I a_Ix_1^{j_1}\cdots j_ix_i^{j_i-1}\cdots x_n^{j_n}.$$
The left $A$-module $Der_k(A)$ is a free $A$-module on the elements $\partial_{x_i}$: There is an isomorphism
$$Der_k(A) \cong A\{\partial_{x_1}, \ldots, \partial_{x_n}\}.$$