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Let $J$ be a Jacobi field along the geodesic $\gamma : [0,a] \to M$. Then $$\langle J(t),\gamma'(t) \rangle=\langle J'(0),\gamma'(0)\rangle t + \langle J(0),\gamma'(0)\rangle,$$ where $t \in [0,a]$.

Suppose that $J(0)=0$. Then $\langle J'(0),\gamma'(0)\rangle = 0$ if and only if $\langle J,\gamma'\rangle \equiv 0$; in particular, the space of Jacobi fields $J$ with $J(0)=0$ and $\langle J,\gamma' \rangle(t) \equiv 0$ has dimension equal to $n-1$.

I get that $J(0)=0$ implies that $\langle J(t),\gamma'(t) \rangle=\langle J'(0),\gamma'(0)\rangle t$, and from this the statement "$\langle J'(0),\gamma'(0)\rangle = 0$ if and only if $\langle J,\gamma'\rangle \equiv 0$" follows.

However, how do we deduce that the dimension of the space of Jacobi fields is $n-1$?

Addendum: do Carmo states also that:

A Jacobi field is determined by its initial conditions $J(0)$, $\frac{DJ}{dt}(0)$. Indeed, let $e_1(t),\ldots,e_n(t)$ be parallel, orthonormal fields along $\gamma$. We shall write: $$J(t)=\sum_i f_i(t) e_i(t), \qquad a_{ij} = \langle R(\gamma'(t),e_i(t))\gamma'(t),e_j(t) \rangle,$$ $i,j=1,\ldots,n=\dim M$. Then $$ \frac{D^2 J}{dt^2} = \sum_i f_i''(t) e_i(t), $$ and \begin{align} R(\gamma',J)\gamma' &= \sum_j \langle R(\gamma',J)\gamma',e_j \rangle e_j \\ &= \sum_{ij} f_i \langle R(\gamma',e_i)\gamma',e_j \rangle e_j \\ &= \sum_{ij} f_i a_{ij} e_j. \end{align} Therefore, the Jacobi equation $\frac{D^2J}{dt^2}+R(\gamma'(t),J(t))\gamma'(t)=0$ is equivalent to the system $$f_j''(t)+\sum_i a_{ij}(t) f_i(t)=0,$$ $j=1,\ldots,n$, which is a linear system of the second order. Hence, given the initial conditions $J(0)=0$, $J'(0)=\frac{DJ}{dt}(0)$, there exists a $C^\infty$ solution of the system, defined on $[0,a]$. There exist, therefore, $2n$ linearly independent Jacobi fields along $\gamma$.

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    Do you know the theorem about existence and uniqueness of Jacobi fields given arbitrary values for $J(0)$ and $J'(0)$? – 2017-02-25
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    @JackLee I've read the discussion of existence and uniqueness of Jacobi fields on page 113 of "Riemannian Geometry" by Manfredo do Carmo. – 2017-02-25
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    OK, so what's the dimension of the space of all Jacobi fields along $\gamma$? And how much is the dimension reduced by the condition $J(0)=0$? And how much more is it reduced by $\langle J'(0),\gamma'(0)\rangle=0$? – 2017-02-25
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    That's not right: you can specify $J(0)$ and $J'(0)$ arbitrarily, so the dimension of the space of Jacobi fields is $2n$. If you write the equation $J(0)=0$ in terms of a basis, you'll see that it is $n$ independent equations. – 2017-02-25
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    No, I just meant choose a basis for $T_{\gamma(0)}M$ and think about what the equation $J(0)=0$ means when expressed in terms of that basis. – 2017-02-26
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    @JackLee I will answer your questions more carefully this time. "OK, so what's the dimension of the space of all Jacobi fields along $\gamma$?" $2n$. "And how much is the dimension reduced by the condition $J(0)=0$?" $n$. "And how much more is it reduced by $\langle J′(0),\gamma′(0) \rangle$?" $1$. So $2n-n-1=n-1$. I realized that the space of Jacobi fields with $J(0)=0$ is $n$ by choosing (as you said) a basis for $T_{\gamma(0)}M$, which has dimension $n$. Furthermore, this is true if and only if $J_1,\ldots,J_n$ with $J_i(0)=0$ are linearly independent. (And I proved it for myself, too.) – 2017-04-17

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If $J$ is a Jacobi field along a geodesic $c(t):=\exp_p\ tv,\ |v|=1$, then assume that $J(0)=0$. Then $$ J(t)=(d\exp_p)_{tv}\ tw$$ s.t. $J'(0)=w$.

If $w=v$, then $J(t)=t c'(t)$. If $(J,c'(t))=0$, then $w\perp v$ (cf. Gauss Lemma). So for choice of $w$ we have dimension $n-1$.